Complexnumbers and polynomials - Haese Mathematics · PDF file... we can add, subtract,...

6Chapter

Contents:

Syllabus reference: 1.5, 1.8, 2.5, 2.6

Complex numbers

A Real quadratics with ¢ < 0B Complex numbers

C Real polynomials

D Zeros, roots, and factors

E Polynomial theorems

F Graphing real polynomials

and polynomials

IB_HL-3edmagentacyan yellow black

0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\173IB_HL-3ed_06.cdr Thursday, 12 April 2012 12:24:44 PM BEN

174 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)

In Chapter 1, we determined that:

If ax2 + bx + c = 0, a 6= 0 and a, b, c 2 R , then the solutions or roots are found using the

formula x =¡b § p

¢

2awhere ¢ = b2 ¡ 4ac is known as the discriminant.

We also observed that if: ² ¢ > 0 we have two real distinct solutions

² ¢ = 0 we have two real identical solutions

² ¢ < 0 we have no real solutions.

However, it is in fact possible to write down two solutions for the case where ¢ < 0. To do this we

need imaginary numbers.

In 1572, Rafael Bombelli defined the imaginary number i =p¡1. It is called ‘imaginary’ because we

cannot place it on a number line. With i defined, we can write down solutions for quadratic equations

with ¢ < 0. They are called complex solutions because they include a real and an imaginary part.

Any number of the form a + bi where a and b are real and i =p¡1,

is called a complex number.

Solve the quadratic equations: a x2 = ¡4 b z2 + z + 2 = 0

a x2 = ¡4

) x = §p¡4

) x = §p

4p¡1

) x = §2i

b z2 + z + 2 has a = 1, b = 1, c = 2

) z =¡1§

p12 ¡ 4(1)(2)

2(1)

) z =¡1§p¡7

2

) z =¡1§ i

p7

2= ¡1

2§

p7

2i

In Example 1 above, notice that ¢ < 0 in both cases. In each case we have found two complex

solutions of the form a + bi, where a and b are real.

Write as a product of linear factors:

a x2 + 4 b x2 + 11

a x2 + 4

= x2 ¡ 4i2

= (x + 2i)(x ¡ 2i)

b x2 + 11

= x2 ¡ 11i2

= (x + ip

11)(x ¡ ip

11)

REAL QUADRATICS WITH ¢ < 0A

Example 2 Self Tutor

Example 1 Self Tutor If the coefficient of

is a square root, we

write the first.

i

i


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\174IB_HL-3ed_06.cdr Thursday, 29 March 2012 9:56:20 AM BEN

COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 175

Solve for x:

a x2 + 9 = 0 b x3 + 2x = 0

a x2 + 9 = 0

) x2 ¡ 9i2 = 0

) (x + 3i)(x ¡ 3i) = 0

) x = §3i

b x3 + 2x = 0

) x(x2 + 2) = 0

) x(x2 ¡ 2i2) = 0

) x(x + ip

2)(x ¡ ip

2) = 0

) x = 0 or §ip

2

Solve for x:

a x2 ¡ 4x + 13 = 0 b x4 + x2 = 6

a x2 ¡ 4x + 13 = 0

) x =4§

p16¡ 4(1)(13)

2

) x =4§p¡36

2

) x =4§ 6i

2

) x = 2 + 3i or 2 ¡ 3i

b x4 + x2 = 6

) x4 + x2 ¡ 6 = 0

) (x2 + 3)(x2 ¡ 2) = 0

) (x + ip

3)(x ¡ ip

3)(x +p

2)(x ¡p

2) = 0

) x = §ip

3 or §p

2

EXERCISE 6A

1 Write in terms of i:

ap¡9 b

p¡64 c

q¡1

4d

p¡5 ep¡8

2 Write as a product of linear factors:

a x2 ¡ 9 b x2 + 9 c x2 ¡ 7 d x2 + 7

e 4x2 ¡ 1 f 4x2 + 1 g 2x2 ¡ 9 h 2x2 + 9

i x3 ¡ x j x3 + x k x4 ¡ 1 l x4 ¡ 16

3 Solve for x:

a x2 ¡ 25 = 0 b x2 + 25 = 0 c x2 ¡ 5 = 0 d x2 + 5 = 0

e 4x2 ¡ 9 = 0 f 4x2 + 9 = 0 g x3 ¡ 4x = 0 h x3 + 4x = 0

i x3 ¡ 3x = 0 j x3 + 3x = 0 k x4 ¡ 1 = 0 l x4 = 81

4 Solve for x:

a x2 ¡ 10x + 29 = 0 b x2 + 6x + 25 = 0 c x2 + 14x + 50 = 0

d 2x2 + 5 = 6x e x2 ¡ 2p

3x + 4 = 0 f 2x +1

x= 1

5 Solve for x:

a x4 + 2x2 = 3 b x4 = x2 + 6 c x4 + 5x2 = 36

d x4 + 9x2 + 14 = 0 e x4 + 1 = 2x2 f x4 + 2x2 + 1 = 0




0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Any number of the form a + bi where a, b 2 R and i =p¡1, is called a complex number.

Notice that all real numbers are complex numbers in the special case where b = 0.

A complex number of the form bi where b 2 R , b 6= 0, is called an imaginary number or

purely imaginary.

THE ‘SUM OF TWO SQUARES’

Notice that a2 + b2 = a2 ¡ b2i2 fas i2 = ¡1g= (a + bi)(a ¡ bi)

Compare: a2 ¡ b2 = (a + b)(a ¡ b) fthe difference of two squares factorisationga2 + b2 = (a + bi)(a ¡ bi) fthe sum of two squares factorisationg

If we write z = a + bi where a, b 2 R , then:

² a is the real part of z and we write a = Re (z)

² b is the imaginary part of z and we write b = Im (z).

For example: If z = 2 + 3i, then Re (z) = 2 and Im (z) = 3.

If z = ¡p

2i, then Re (z) = 0 and Im (z) = ¡p

2.

OPERATIONS WITH COMPLEX NUMBERS

Operations with complex numbers are identical to those with radicals, but with i2 = ¡1 rather than

(p

2)2 = 2 or (p

3)2 = 3.

For example:

² addition: (2 +p

3) + (4 + 2p

3) = (2 + 4) + (1 + 2)p

3 = 6 + 3p

3

(2 + i) + (4 + 2i) = (2 + 4) + (1 + 2)i = 6 + 3i

² multiplication: (2 +p

3)(4 + 2p

3) = 8 + 4p

3 + 4p

3 + 2(p

3)2 = 8 + 8p

3 + 6

(2 + i)(4 + 2i) = 8 + 4i + 4i + 2i2 = 8 + 8i ¡ 2

So, we can add, subtract, multiply, and divide complex numbers in the same way we perform these

operations with radicals:

(a + bi) + (c + di) = (a + c) + (b + d)i addition

(a + bi) ¡ (c + di) = (a ¡ c) + (b ¡ d)i subtraction

(a + bi)(c + di) = ac + adi + bci + bdi2 multiplication

a + bi

c + di=

µa + bi

c + di

¶µc ¡ di

c ¡ di

¶=

ac ¡ adi + bci ¡ bdi2

c2 + d2division

Notice that in the division process, we use a multiplication technique to obtain a real number in the

denominator.

COMPLEX NUMBERSB


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



If z = 3 + 2i and w = 4 ¡ i find:

a z + w b z ¡ w c zw dz

w

a z + w

= (3 + 2i) + (4 ¡ i)

= 7 + i

b z ¡ w

= (3 + 2i) ¡ (4 ¡ i)

= 3 + 2i ¡ 4 + i

= ¡1 + 3i

c zw

= (3 + 2i)(4 ¡ i)

= 12 ¡ 3i + 8i ¡ 2i2

= 12 + 5i + 2

= 14 + 5i

dz

w=

3 + 2i

4¡ i

=³3 + 2i

4¡ i

´³4 + i

4 + i

´

=12 + 3i+ 8i+ 2i2

16¡ i2

=10 + 11i

17

= 10

17+ 11

17i

You can use your calculator to perform operations with complex numbers.

EXERCISE 6B.1

1 Copy and complete: z Re (z) Im (z)

3 + 2i

5 ¡ i

3

0

z Re (z) Im (z)

¡3 + 4i

¡7 ¡ 2i

¡11i

ip

3

2 If z = 5 ¡ 2i and w = 2 + i, find in simplest form:

a z + w b 2z c iw d z ¡ w

e 2z ¡ 3w f zw g w2 h z2

3 For z = 1 + i and w = ¡2 + 3i, find in simplest form:

a z + 2w b z2 c z3 d iz

e w2 f zw g z2w h izw

4 a Simplify in for n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and also for n = ¡1, ¡2, ¡3, ¡4, ¡5.

b Hence, simplify i4n+3 where n is any integer.

5 Write (1 + i)4 in simplest form. Hence, find (1 + i)101 in simplest form.

6 Suppose z = 2 ¡ i and w = 1 + 3i. Write in exact form a + bi where a, b 2 R :

az

wb

i

zc

w

izd z¡2


After solving the questions in the following exercise by hand, check your

answers using technology. GRAPHICS

CALCULATOR

INSTRUCTIONS


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



7 Simplify: ai

1¡ 2ib

i(2¡ i)

3¡ 2ic

1

2¡ i¡ 2

2 + i

8 If z = 2 + i and w = ¡1 + 2i, find:

a Im (4z ¡ 3w) b Re (zw) c Im (iz2) d Re

³z

w

´

EQUALITY OF COMPLEX NUMBERS

Two complex numbers are equal when their real parts are equal and their imaginary parts are equal.

a + bi = c + di , a = c and b = d.

Proof: Suppose b 6= d. Now if a + bi = c + di where a, b, c, and d are real,

then bi ¡ di = c ¡ a

) i(b ¡ d) = c ¡ a

) i =c¡ a

b¡ dfas b 6= dg

This is false as the RHS is real but the LHS is imaginary.

Thus, the supposition is false. Hence b = d and furthermore a = c.

For the complex number a + bi, where a and b are real, a + bi = 0 , a = 0 and b = 0.

Find real numbers x and y such that:

a (x + yi)(2 ¡ i) = ¡i b (x + 2i)(1 ¡ i) = 5 + yi

a (x + yi)(2 ¡ i) = ¡i

) x + yi =¡i

2¡ i

=³ ¡i

2¡ i

´³2 + i

2 + i

´

=¡2i¡ i2

4 + 1

=1¡ 2i

5

= 1

5¡ 2

5i

Equating real and imaginary parts, x = 1

5and y = ¡2

5.

b (x + 2i)(1 ¡ i) = 5 + yi

) x ¡ xi + 2i + 2 = 5 + yi

) (x + 2) + (2 ¡ x)i = 5 + yi

Equating real and imaginary parts,

) x + 2 = 5 and 2 ¡ x = y

) x = 3 and y = ¡1

EXERCISE 6B.2

1 Find exact real numbers x and y such that:

a 2x + 3yi = ¡x ¡ 6i b x2 + xi = 4 ¡ 2i

c (x + yi)(2 ¡ i) = 8 + i d (3 + 2i)(x + yi) = ¡i

2 Find exact x, y 2 R such that:

a 2(x + yi) = x ¡ yi b (x + 2i)(y ¡ i) = ¡4 ¡ 7i

c (x + i)(3 ¡ iy) = 1 + 13i d (x + yi)(2 + i) = 2x ¡ (y + 1)i



0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


HISTORICAL NOTE


3 The complex number z satisfies the equation 3z + 17i = iz + 11. Write z in the form a + bi

where a, b 2 R and i =p¡1.

18th century mathematicians enjoyed playing with these new ‘imaginary’

numbers, but they were regarded as little more than interesting curiosities

until the work of Gauss (1777 - 1855), the German mathematician,

astronomer, and physicist.

For centuries mathematicians had attempted to find a method of trisecting

an angle using a compass and straight edge. Gauss put an end to this

when he used complex numbers to prove the impossibility of such a

construction. By his systematic use of complex numbers, he was able to

convince mathematicians of their usefulness.

Early last century, the American engineer Steinmetz used complex numbers to solve electrical

problems, illustrating that complex numbers did have a practical application.

Complex numbers are now used extensively in electronics, engineering, and physics.

COMPLEX CONJUGATES

Complex numbers a + bi and a ¡ bi are called complex conjugates.

If z = a + bi we write its conjugate as z¤ = a ¡ bi.

We saw on page 176 that the complex conjugate is important for division:

z

w=

z

w

w¤

w¤ =zw¤

ww¤ which makes the denominator real.

Quadratics with real coefficients are called real quadratics. This does not necessarily

mean that their zeros are real.

² If a quadratic equation has rational coefficients and an irrational root of the form

c + dpn, then the conjugate c ¡ d

pn is also a root of the quadratic equation.

² If a real quadratic equation has ¢ < 0 and c + di is a complex root, then the

complex conjugate c ¡ di is also a root.

For example: ² x2 ¡ 2x + 5 = 0 has ¢ = (¡2)2 ¡ 4(1)(5) = ¡16

and the solutions are x = 1 + 2i and 1 ¡ 2i

² x2 + 4 = 0 has ¢ = 02 ¡ 4(1)(4) = ¡16

and the solutions are x = 2i and ¡2i

Carl Friedrich Gauss

4 Express z in the form a + bi where a, b 2 Z , if z =³

4

1 + i+ 7 ¡ 2i

´2.

5 Find the real values of m and n for which 3(m + ni) = n ¡ 2mi ¡ (1 ¡ 2i).

6 Express z =3ip2¡ i

+ 1 in the form a + bi where a, b 2 R are given exactly.

7 Suppose (a + bi)2 = ¡16 ¡ 30i where a, b 2 R and a > 0.

Find the possible values of a and b.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\179IB_HL-3ed_06.cdr Monday, 23 April 2012 3:34:02 PM BEN


Theorem: If c + di and c ¡ di are roots of a quadratic equation, then the quadratic equation

is a(x2 ¡ 2cx + (c2 + d2)) = 0 for some constant a 6= 0.

Proof: The sum of the roots = 2c and the product = (c + di)(c ¡ di) = c2 + d2

) x2 ¡ (sum)x + (product) = 0

) x2 ¡ 2cx + (c2 + d2) = 0

In general, a(x2 ¡ 2cx + c2 + d2) = 0 for some constant a 6= 0.

Notice that: ² the sum of complex conjugates c + di and c ¡ di is 2c which is real

² the product is (c + di)(c ¡ di) = c2 + d2 which is also real.

Find all real quadratic equations having 1 ¡ 2i as a root.

As 1 ¡ 2i is a root, 1 + 2i is also a root.

The sum of the roots = 1 ¡ 2i + 1 + 2i

= 2

The product of the roots = (1 ¡ 2i)(1 + 2i)

= 1 + 4

= 5

So, as x2 ¡ (sum)x + (product) = 0,

a(x2 ¡ 2x + 5) = 0, a 6= 0 gives all possible equations.

Find exact values for a and b ifp

2 + i is a root of x2 + ax + b = 0, a, b 2 R .

Since a and b are real, the quadratic has real coefficients.

)p

2 ¡ i is also a root.

The sum of the roots =p

2 + i +p

2 ¡ i = 2p

2.

The product of the roots = (p

2 + i)(p

2 ¡ i) = 2 + 1 = 3.

Thus a = ¡2p

2 and b = 3.

EXERCISE 6B.3

1 Find all quadratic equations with real coefficients and roots of:

a 3 § i b 1 § 3i c ¡2 § 5i dp

2 § i

e 2 § p3 f 0 and ¡2

3g §i

p2 h ¡6 § i

2 Find exact values for a and b if:

a 3 + i is a root of x2 + ax + b = 0, where a and b are real

b 1 ¡ p2 is a root of x2 + ax + b = 0, where a and b are rational

c a + ai is a root of x2 + 4x + b = 0, where a and b are real. [Careful!]




0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


PROPERTIES OF CONJUGATESINVESTIGATION 1


PROPERTIES OF CONJUGATES

The purpose of this investigation is to discover any properties that complex conjugates might have.

What to do:

1 Given z1 = 1 ¡ i and z2 = 2 + i find:

a z ¤1 b z ¤

2 c (z ¤1 )¤ d (z ¤

2 )¤

e (z1 + z2)¤ f z ¤

1 + z ¤2 g (z1 ¡ z2)

¤ h z ¤1 ¡ z ¤

2

i (z1z2)¤ j z ¤

1 z ¤2 k

³z1

z2

´¤l

z ¤1

z ¤2

m (z 21 )¤ n (z ¤

1 )2 o (z 32 )¤ p (z ¤

2 )3

2 Repeat 1 with z1 and z2 of your choice.

3 Examine your results from 1 and 2, and hence suggest some rules for complex conjugates.

From the Investigation you should have discovered the following rules for complex conjugates:

² (z¤)¤ = z

² (z1 + z2)¤ = z ¤

1 + z ¤2 and (z1 ¡ z2)

¤ = z ¤1 ¡ z ¤

2

² (z1z2)¤ = z ¤

1 £ z ¤2 and

µz1

z2

¶¤=

z ¤1

z ¤2

, z2 6= 0

² (zn)¤ = (z¤)n for positive integers n

² z + z¤ and zz¤ are real.

Show that for all complex numbers z1 and z2:

a (z1 + z2)¤ = z ¤

1 + z ¤2 b (z1z2)

¤ = z ¤1 £ z ¤

2

a Let z1 = a + bi

) z ¤1 = a ¡ bi

and

and

z2 = c + di

z ¤2 = c ¡ di

Now z1 + z2 = (a + c) + (b + d)i

) (z1 + z2)¤ = (a + c) ¡ (b + d)i

= a + c ¡ bi ¡ di

= a ¡ bi + c ¡ di

= z ¤1 + z ¤

2

b Let z1 = a + bi and z2 = c + di

) z1z2 = (a + bi)(c + di)

= ac + adi + bci + bdi2

= (ac ¡ bd) + i(ad + bc)

) (z1z2)¤ = (ac ¡ bd) ¡ i(ad + bc) .... (1)

Also, z ¤1 £ z ¤

2

= (a ¡ bi)(c ¡ di)

= ac ¡ adi ¡ bci + bdi2

= (ac ¡ bd) ¡ i(ad + bc) .... (2)

From (1) and (2), (z1z2)¤ = z ¤

1 £ z ¤2



0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



CONJUGATE GENERALISATIONS

Notice that (z1 + z2 + z3)¤ = (z1 + z2)

¤ + z ¤3 ftreating z1 + z2 as one complex numberg

= z ¤1 + z ¤

2 + z ¤3 .... (1)

Likewise (z1 + z2 + z3 + z4)¤ = (z1 + z2 + z3)

¤ + z ¤4

= z ¤1 + z ¤

2 + z ¤3 + z ¤

4 ffrom (1)gThere is no reason why this process cannot continue for the conjugate of 5, 6, 7, .... complex numbers.

We can therefore generalise the result:

(z1 + z2 + z3 + :::: + zn)¤ = z ¤1 + z ¤

2 + z ¤3 + :::: + z ¤

n

The process of proving the general case using the simpler cases when n = 1, 2, 3, 4, .... requires

mathematical induction. Formal proof by the Principle of Mathematical Induction is discussed in

Chapter 9.

EXERCISE 6B.4

1 Show that (z1 ¡ z2)¤ = z ¤

1 ¡ z ¤2 for all complex numbers z1 and z2.

2 Simplify the expression (w¤ ¡ z)¤ ¡ (w ¡ 2z¤) using the properties of conjugates.

3 It is known that a complex number z satisfies the equation z¤ = ¡z.

Show that z is either purely imaginary or zero.

4 Suppose z1 = a + bi and z2 = c + di are complex numbers.

a Findz1

z2in the form X + iY . b Show that

³z1

z2

´¤=

z ¤1

z ¤2

for all z1 and z2 6= 0.

5 a An easier way of proving

³z1

z2

´¤=

z ¤1

z ¤2

is to start with

³z1

z2

´¤£ z ¤

2 .

Show how this can be done, remembering we have already proved that “the conjugate of a

product is the product of the conjugates” in Example 9.

b Let z = a + bi be a complex number. Prove the following:

i If z = z¤, then z is real. ii If z¤ = ¡z, then z is purely imaginary or zero.

6 Prove that for all complex numbers z and w:

a zw¤ + z¤w is always real b zw¤ ¡ z¤w is purely imaginary or zero.

7 a If z = a + bi, find z2 in the form X + iY .

b Hence, show that (z2)¤ = (z¤)2 for all complex numbers z.

c Repeat a and b but for z3 instead of z2.

8 Suppose w =z ¡ 1

z¤ + 1where z = a + bi. Find the conditions under which:

a w is real b w is purely imaginary.

9 a Assuming (z1z2)¤ = z ¤

1 z ¤2 , explain why (z1z2z3)

¤ = z ¤1 z ¤

2 z ¤3 .

b Hence show that (z1z2z3z4)¤ = z ¤

1 z ¤2 z ¤

3 z ¤4 .

c Generalise your results from a and b.

d Given your generalisation in c, what is the result of letting all zi values be equal to z?


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Up to this point we have studied linear and quadratic functions at some depth, with perhaps occasional

reference to cubic functions. These are part of a larger family of functions called the polynomials.

A polynomial function is a function of the form

P (x) = anxn + an¡1x

n¡1 + :::: + a2x2 + a1x + a0, a1, ...., an constant, an 6= 0.

We say that: x is the variable

a0 is the constant term

an is the leading coefficient and is non-zero

ar is the coefficient of xr for r = 0, 1, 2, ...., n

n is the degree of the polynomial, being the highest power of the variable.

In summation notation, we write P (x) =nP

r=0

arxr,

which reads: “the sum from r = 0 to n, of arxr ”.

A real polynomial P (x) is a polynomial for which ar 2 R , r = 0, 1, 2, ...., n.

The low degree members of the polynomial family have special names, some of which you are already

familiar with. For these polynomials, we commonly write their coefficients as a, b, c, ....

Polynomial function Degree Name

ax + b, a 6= 0 1 linear

ax2 + bx + c, a 6= 0 2 quadratic

ax3 + bx2 + cx + d, a 6= 0 3 cubic

ax4 + bx3 + cx2 + dx + e, a 6= 0 4 quartic

ADDITION AND SUBTRACTION

To add or subtract two polynomials, we collect ‘like’ terms.

If P (x) = x3 ¡ 2x2 + 3x ¡ 5 and Q(x) = 2x3 + x2 ¡ 11, find:

a P (x) + Q(x) b P (x) ¡ Q(x)

a P (x) + Q(x)

= x3 ¡ 2x2 + 3x ¡ 5

+ 2x3 + x2 ¡ 11

= 3x3 ¡ x2 + 3x ¡ 16

b P (x) ¡ Q(x)

= x3 ¡ 2x2 + 3x ¡ 5 ¡ (2x3 + x2 ¡ 11)

= x3 ¡ 2x2 + 3x ¡ 5

¡ 2x3 ¡ x2 + 11

= ¡x3 ¡ 3x2 + 3x + 6

REAL POLYNOMIALSC


Collecting ‘like’ terms is

made easier by writing them

one above the other.

It is a good idea to place

brackets around expressions

which are subtracted.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



SCALAR MULTIPLICATION

To multiply a polynomial by a scalar (constant) we multiply each term by the scalar.

If P (x) = x4 ¡ 2x3 + 4x + 7 find: a 3P (x) b ¡2P (x)

a 3P (x)

= 3(x4 ¡ 2x3 + 4x + 7)

= 3x4 ¡ 6x3 + 12x + 21

b ¡2P (x)

= ¡2(x4 ¡ 2x3 + 4x + 7)

= ¡2x4 + 4x3 ¡ 8x ¡ 14

POLYNOMIAL MULTIPLICATION

To multiply two polynomials, we multiply each term of the first polynomial by each

term of the second polynomial, and then collect like terms.

If P (x) = x3 ¡ 2x + 4 and Q(x) = 2x2 + 3x ¡ 5, find P (x)Q(x).

P (x)Q(x) = (x3 ¡ 2x + 4)(2x2 + 3x ¡ 5)

= x3(2x2 + 3x ¡ 5) ¡ 2x(2x2 + 3x ¡ 5) + 4(2x2 + 3x ¡ 5)

= 2x5 + 3x4 ¡ 5x3

¡ 4x3 ¡ 6x2 + 10x

+ 8x2 + 12x ¡ 20

= 2x5 + 3x4 ¡ 9x3 + 2x2 + 22x ¡ 20

SYNTHETIC MULTIPLICATION (OPTIONAL)

Polynomial multiplication can be performed using the coefficients only. We call this synthetic

multiplication.

For example, for (x3 + 2x ¡ 5)(2x + 3) we detach coefficients and multiply. It is different from the

ordinary multiplication of large numbers because we sometimes have negative coefficients, and because

we do not carry tens into the next column.



1 0 2 ¡5

£ 2 3

3 0 6 ¡15

2 0 4 ¡10

2 3 4 ¡4 ¡15

x4 x3 x2 x constants

coefficients of x3 + 2x ¡ 5

coefficients of 2x + 3

So (x3 + 2x ¡ 5)(2x + 3)

= 2x4 + 3x3 + 4x2 ¡ 4x ¡ 15.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



EXERCISE 6C.1

1 If P (x) = x2 + 2x + 3 and Q(x) = 4x2 + 5x + 6, find in simplest form:

a 3P (x) b P (x) + Q(x) c P (x) ¡ 2Q(x) d P (x)Q(x)

2 If f(x) = x2 ¡ x + 2 and g(x) = x3 ¡ 3x + 5, find in simplest form:

a f(x) + g(x) b g(x) ¡ f(x) c 2f(x) + 3g(x)

d g(x) + xf(x) e f(x) g(x) f [f(x)]2

3 Expand and simplify:

a (x2 ¡ 2x + 3)(2x + 1) b (x ¡ 1)2(x2 + 3x ¡ 2) c (x + 2)3

d (2x2 ¡ x + 3)2 e (2x ¡ 1)4 f (3x ¡ 2)2(2x + 1)(x ¡ 4)

4 Find the following products:

a (2x2 ¡ 3x + 5)(3x ¡ 1) b (4x2 ¡ x + 2)(2x + 5)

c (2x2 + 3x + 2)(5 ¡ x) d (x ¡ 2)2(2x + 1)

e (x2 ¡ 3x + 2)(2x2 + 4x ¡ 1) f (3x2 ¡ x + 2)(5x2 + 2x ¡ 3)

g (x2 ¡ x + 3)2 h (2x2 + x ¡ 4)2

i (2x + 5)3 j (x3 + x2 ¡ 2)2

DIVISION OF POLYNOMIALS

The division of polynomials is only useful if we divide a polynomial of degree n by another of degree n

or less.

DIVISION BY LINEARS

Consider (2x2 + 3x + 4)(x + 2) + 7.

If we expand this expression we obtain (2x2 + 3x + 4)(x + 2) + 7 = 2x3 + 7x2 + 10x + 15.

Dividing both sides by (x + 2), we obtain

2x3 + 7x2 + 10x+ 15

x+ 2=

(2x2 + 3x+ 4)(x+ 2) + 7

x+ 2

=(2x2 + 3x+ 4)(x+ 2)

x+ 2+

7

x+ 2

= 2x2 + 3x + 4 +7

x+ 2where x + 2 is the divisor,

2x2 + 3x + 4 is the quotient,

and 7 is the remainder.

If P (x) is divided by ax + b until a constant remainder R is obtained, then

P (x)

ax + b= Q(x) +

R

ax + bwhere ax + b is the divisor, D(x),

Q(x) is the quotient,

and R is the remainder.

Notice that P (x) = Q(x) £ (ax + b) + R.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



DIVISION ALGORITHM

We can divide a polynomial by another polynomial using an algorithm similar to that used for division

of whole numbers:

The division algorithm can also be performed by leaving

out the variable, as shown alongside.

Either way,2x3 + 7x2 + 10x+ 15

x+ 2= 2x2 + 3x + 4 +

7

x+ 2

Find the quotient and remainder forx3 ¡ x2 ¡ 3x¡ 5

x¡ 3.

Hence write x3 ¡ x2 ¡ 3x ¡ 5 in the form Q(x) £ (x ¡ 3) + R.

The quotient is x2 + 2x + 3and the remainder is 4.

)x3 ¡ x2 ¡ 3x¡ 5

x¡ 3= x2 + 2x + 3 +

4

x¡ 3

) x3 ¡ x2 ¡ 3x ¡ 5 = (x2 + 2x + 3)(x ¡ 3) + 4.


Step 1: What do we multiply x by to get 2x3?

The answer is 2x2,

and 2x2(x + 2) = 2x3 + 4x2| {z }.Step 2: Subtract 2x3 + 4x2 from 2x3 + 7x2.

The answer is 3x2.

Step 3: Bring down the 10x to obtain 3x2 + 10x.

Return to Step 1 with the question:

“What must we multiply x by to get 3x2?”

The answer is 3x, and 3x(x + 2) = 3x2 + 6x ....

We continue the process until we are left with a constant.

2x2 + 3x + 4

x + 2 2x3 + 7x2 + 10x + 15

¡ (2x3 + 4x2)

3x2 + 10x

¡ (3x2 + 6x)

4x + 15

¡ (4x + 8)

7

2 3 4

1 2 2 7 10 15

¡ (2 4)

3 10

¡ (3 6)

4 15

¡ (4 8)

7

Check your answer by

expanding the RHS.

x2 + 2x + 3

x ¡ 3 x3 ¡ x2 ¡ 3x ¡ 5

2x2 ¡ 3x

3x ¡ 5

4

¡ (x3¡ 3x2)

¡ (2x2¡ 6x)

¡ (3x ¡ 9)


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



EXERCISE 6C.2

1 Find the quotient and remainder for the following, and hence write the division in the form

P (x) = Q(x)D(x) + R, where D(x) is the divisor.

ax2 + 2x¡ 3

x+ 2b

x2 ¡ 5x+ 1

x¡ 1c

2x3 + 6x2 ¡ 4x+ 3

x¡ 2

2 Perform the following divisions, and hence write the division in the form

P (x) = Q(x)D(x) + R.

ax2 ¡ 3x+ 6

x¡ 4b

x2 + 4x¡ 11

x+ 3c

2x2 ¡ 7x+ 2

x¡ 2

d2x3 + 3x2 ¡ 3x¡ 2

2x+ 1e

3x3 + 11x2 + 8x+ 7

3x¡ 1f

2x4 ¡ x3 ¡ x2 + 7x+ 4

2x+ 3

3 Perform the divisions:

ax2 + 5

x¡ 2b

2x2 + 3x

x+ 1c

3x2 + 2x¡ 5

x+ 2

dx3 + 2x2 ¡ 5x+ 2

x¡ 1e

2x3 ¡ x

x+ 4f

x3 + x2 ¡ 5

x¡ 2

SYNTHETIC DIVISION (OPTIONAL)

Click on the icon for an exercise involving a synthetic division process for the division

of a polynomial by a linear.

Perform the divisionx4 + 2x2 ¡ 1

x+ 3.

Hence write x4 + 2x2 ¡ 1 in the form Q(x) £ (x + 3) + R.

Notice the insertion

of 0x3 and 0x.x3 ¡ 3x2 + 11x ¡ 33

x + 3 x4 + 0x3 + 2x2 + 0x ¡ 1

¡ (x4 + 3x3)

¡ 3x3 + 2x2

¡ (¡ 3x3 ¡ 9x2)

11x2 + 0x

¡ (11x2 + 33x)

¡ 33x ¡ 1

¡ (¡ 33x ¡ 99)

98

)x4 + 2x2 ¡ 1

x+ 3= x3 ¡ 3x2 + 11x ¡ 33 +

98

x+ 3

) x4 + 2x2 ¡ 1 = (x3 ¡ 3x2 + 11x ¡ 33)(x + 3) + 98


PRINTABLE

SECTION


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



DIVISION BY QUADRATICS

As with division by linears we can use the division algorithm to divide polynomials by quadratics. The

division process stops when the remainder has degree less than that of the divisor, so

If P (x) is divided by ax2 + bx + c then

P (x)

ax2 + bx + c= Q(x) +

ex + f

ax2 + bx + cwhere ax2 + bx + c is the divisor,

Q(x) is the quotient,

andex + f

ax2 + bx + cis the remainder.

The remainder will be linear if e 6= 0, and constant if e = 0.

Find the quotient and remainder forx4 + 4x3 ¡ x+ 1

x2 ¡ x+ 1.

Hence write x4 + 4x3 ¡ x + 1 in the form Q(x) £ (x2 ¡ x + 1) + R(x).

The quotient is x2 + 5x + 4and the remainder is ¡2x ¡ 3.

) x4 + 4x3 ¡ x + 1

= (x2 + 5x + 4)(x2 ¡ x + 1) ¡ 2x ¡ 3

EXERCISE 6C.3

1 Find the quotient and remainder for:

ax3 + 2x2 + x¡ 3

x2 + x+ 1b

3x2 ¡ x

x2 ¡ 1c

3x3 + x¡ 1

x2 + 1d

x¡ 4

x2 + 2x¡ 1

2 Carry out the following divisions and also write each in the form P (x) = Q(x)D(x) + R(x):

ax2 ¡ x+ 1

x2 + x+ 1b

x3

x2 + 2c

x4 + 3x2 + x¡ 1

x2 ¡ x+ 1

d2x3 ¡ x+ 6

(x¡ 1)2e

x4

(x+ 1)2f

x4 ¡ 2x3 + x+ 5

(x¡ 1)(x+ 2)

3 Suppose P (x) = (x ¡ 2)(x2 + 2x + 3) + 7. Find the quotient and remainder when P (x) is

divided by x ¡ 2.

4 Suppose f(x) = (x¡ 1)(x+2)(x2 ¡ 3x+5)+15 ¡ 10x. Find the quotient and remainder when

f(x) is divided by x2 + x ¡ 2.


x2 + 5x + 4

x2 ¡ x + 1 x4 + 4x3 + 0x2 ¡ x + 1

¡ (x4 ¡ x3 + x2)

5x3 ¡ x2 ¡ x

¡ (5x3 ¡ 5x2 + 5x)

4x2 ¡ 6x + 1

¡ (4x2 ¡ 4x + 4)

¡ 2x ¡ 3


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



A zero of a polynomial is a value of the variable which makes the polynomial equal to zero.

® is a zero of polynomial P (x) , P (®) = 0.

The roots of a polynomial equation are the solutions to the equation.

® is a root (or solution) of P (x) = 0 , P (®) = 0.

The roots of P (x) = 0 are the zeros of P (x) and the x-intercepts of the graph of y = P (x).

Consider P (x) = x3 + 2x2 ¡ 3x ¡ 10

) P (2) = 23 + 2(2)2 ¡ 3(2) ¡ 10

= 8 + 8 ¡ 6 ¡ 10

= 0

An equation has roots.

A polynomial has zeros.

This tells us: ² 2 is a zero of x3 + 2x2 ¡ 3x ¡ 10

² 2 is a root of x3 + 2x2 ¡ 3x ¡ 10 = 0

² the graph of y = x3 + 2x2 ¡ 3x ¡ 10 has the x-intercept 2.

If P (x) = (x + 1)(2x ¡ 1)(x + 2), then (x + 1), (2x ¡ 1), and (x + 2) are its linear factors.

Likewise P (x) = (x+3)2(2x+3) has been factorised into 3 linear factors, one of which is repeated.

(x ¡ ®) is a factor of the polynomial P (x) , there exists a polynomial Q(x)

such that P (x) = (x ¡ ®)Q(x).

Find the zeros of: a x2 ¡ 4x + 53 b z3 + 3z

a We wish to find x such that

x2 ¡ 4x + 53 = 0

) x =4§

p16¡ 4(1)(53)

2

) x =4§p¡196

2=

4§ 14i

2

) x = 2 § 7i

The zeros are 2 ¡ 7i and 2 + 7i.

b We wish to find z such that

z3 + 3z = 0

) z(z2 + 3) = 0

) z(z + ip

3)(z ¡ ip

3) = 0

) z = 0 or § ip

3

The zeros are ¡ip

3, 0, and ip

3.

EXERCISE 6D.1

1 Find the zeros of:

a 2x2 ¡ 5x ¡ 12 b x2 + 6x + 10 c z2 ¡ 6z + 6

d x3 ¡ 4x e z3 + 2z f z4 + 4z2 ¡ 5

2 Find the roots of:

a 5x2 = 3x + 2 b (2x + 1)(x2 + 3) = 0 c ¡2z(z2 ¡ 2z + 2) = 0

d x3 = 5x e z3 + 5z = 0 f z4 = 3z2 + 10

ZEROS, ROOTS, AND FACTORSD



0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Factorise: a 2x3 + 5x2 ¡ 3x b z2 + 4z + 9

a 2x3 + 5x2 ¡ 3x

= x(2x2 + 5x ¡ 3)

= x(2x ¡ 1)(x + 3)

b z2 + 4z + 9 is zero when z =¡4§

p16¡ 4(1)(9)

2

) z =¡4§p¡20

2

) z =¡4§ 2i

p5

2

) z = ¡2 § ip

5

) z2 + 4z + 9 = (z ¡ [¡2 + ip

5])(z ¡ [¡2 ¡ ip

5])

= (z + 2 ¡ ip

5)(z + 2 + ip

5)

3 Find the linear factors of:

a 2x2 ¡ 7x ¡ 15 b z2 ¡ 6z + 16 c x3 + 2x2 ¡ 4x

d 6z3 ¡ z2 ¡ 2z e z4 ¡ 6z2 + 5 f z4 ¡ z2 ¡ 2

4 If P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °) then ®, ¯, and ° are its zeros.

Verify this statement by finding P (®), P (¯), and P (°).

Find all cubic polynomials with zeros 1

2and ¡3 § 2i.

The zeros ¡3 § 2i have sum = ¡3 + 2i ¡ 3 ¡ 2i = ¡6 and

product = (¡3 + 2i)(¡3 ¡ 2i) = 13

) they come from the quadratic factor x2 + 6x + 131

2comes from the linear factor 2x ¡ 1.

) P (x) = a(2x ¡ 1)(x2 + 6x + 13), a 6= 0.

Find all quartic polynomials with zeros 2, ¡1

3, and ¡1 § p

5.

The zeros ¡1 § p5 have sum = ¡1 +

p5 ¡ 1 ¡

p5 = ¡2 and

product = (¡1 +p

5)(¡1 ¡p

5) = ¡4

) they come from the quadratic factor x2 + 2x ¡ 4.

The zeros 2 and ¡1

3come from the linear factors (x ¡ 2) and (3x + 1).

) P (x) = a(x ¡ 2)(3x + 1)(x2 + 2x ¡ 4), a 6= 0.

5 Find all cubic polynomials with zeros:

a §2, 3 b ¡2, §i c 3, ¡1 § i d ¡1, ¡2 § p2

6 Find all quartic polynomials with zeros of:

a §1, §p2 b 2, ¡1, §i

p3 c §p

3, 1 § i d 2 § p5, ¡2 § 3i





0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



POLYNOMIAL EQUALITY

If we know that two polynomials are equal then we can equate coefficients to find unknown coefficients.

For example, if 2x3 + 3x2 ¡ 4x + 6 = ax3 + bx2 + cx + d, where a, b, c, d 2 R , then

a = 2, b = 3, c = ¡4, and d = 6.

Find constants a, b, and c given that:

6x3 + 7x2 ¡ 19x + 7 = (2x ¡ 1)(ax2 + bx + c) for all x.

6x3 + 7x2 ¡ 19x + 7 = (2x ¡ 1)(ax2 + bx + c)

) 6x3 + 7x2 ¡ 19x + 7 = 2ax3 + 2bx2 + 2cx ¡ ax2 ¡ bx ¡ c

) 6x3 + 7x2 ¡ 19x + 7 = 2ax3 + (2b ¡ a)x2 + (2c ¡ b)x ¡ c

Since this is true for all x, we equate coefficients:

Find constants a and b if z4 + 9 = (z2 + az + 3)(z2 + bz + 3) for all z.

z4 + 9 = (z2 + az + 3)(z2 + bz + 3) for all z

) z4 + 9 = z4 + bz3 + 3z2

+ az3 + abz2 + 3az

+ 3z2 + 3bz + 9

) z4 + 9 = z4 + (a + b)z3 + (ab + 6)z2 + (3a + 3b)z + 9 for all z

From (1) and (3) we see that b = ¡a

) in (2), a(¡a) + 6 = 0

) a2 = 6

) a = §p

6 and so b = ¨p

6

) a =p

6, b = ¡p6 or a = ¡p

6, b =p

6



) 2a = 6| {z }x3 s

2b ¡ a = 7| {z }x2 s

2c ¡ b = ¡19| {z }x s

and 7 = ¡c| {z }constants

) a = 3 and c = ¡7 and consequently 2b ¡ 3 = 7 and ¡14 ¡ b = ¡19| {z }) b = 5

in both equations

So, a = 3, b = 5, and c = ¡7.

Equating coefficients gives

8<:

a + b = 0 .... (1) fz3 sgab + 6 = 0 .... (2) fz2 sg

3a + 3b = 0 .... (3) fz sg

When simultaneously

solving more equations

than there are

unknowns, we must

check that any solutions

fit equations. If they

do not, there are

.

all

no

solutions

Two polynomials are if and only if they have the (order) and corresponding

terms have equal coefficients.

equal same degree


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\191IB_HL-3ed_06.cdr Tuesday, 1 May 2012 5:12:59 PM BEN


(x + 3) is a factor of P (x) = x3 + ax2 ¡ 7x + 6. Find a 2 R and the other factors.

Since (x + 3) is a factor, The coefficient of

x3 is 1£ 1 = 1

This must be 2 so the

constant term is 3£ 2 = 6

x3 + ax2 ¡ 7x + 6 = (x + 3)(x2 + bx + 2) for some constant b

= x3 + bx2 + 2x

+ 3x2 + 3bx + 6

= x3 + (b + 3)x2 + (3b + 2)x + 6

Equating coefficients gives 3b + 2 = ¡7 and a = b + 3

) b = ¡3 and a = 0

) P (x) = (x + 3)(x2 ¡ 3x + 2)

= (x + 3)(x ¡ 1)(x ¡ 2)

The other factors are (x ¡ 1) and (x ¡ 2).

(2x + 3) and (x ¡ 1) are factors of 2x4 + ax3 ¡ 3x2 + bx + 3.

Find constants a and b and all zeros of the polynomial.

Since (2x + 3) and (x ¡ 1) are factors, The coefficient of x4

is 2£ 1£ 1 = 2

This must be ¡1 so the constant

term is 3£¡1£¡1 = 3

2x4 + ax3 ¡ 3x2 + bx + 3 = (2x + 3)(x ¡ 1)(x2 + cx ¡ 1) for some c

= (2x2 + x ¡ 3)(x2 + cx ¡ 1)

= 2x4 + 2cx3 ¡ 2x2

+ x3 + cx2 ¡ x

¡ 3x2 ¡ 3cx + 3

= 2x4 + (2c + 1)x3 + (c ¡ 5)x2 + (¡1 ¡ 3c)x + 3

Equating coefficients gives 2c + 1 = a, c ¡ 5 = ¡3, and ¡1 ¡ 3c = b

) c = 2

) a = 5 and b = ¡7

) P (x) = (2x + 3)(x ¡ 1)(x2 + 2x ¡ 1)

Now x2 + 2x ¡ 1 has zeros¡2§

p4¡ 4(1)(¡1)

2=

¡2§ 2p2

2= ¡1 § p

2

) P (x) has zeros ¡3

2, 1, and ¡1 § p

2.




0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



EXERCISE 6D.2

1 Find constants a, b, and c given that:

a 2x2 + 4x + 5 = ax2 + [2b ¡ 6]x + c for all x

b 2x3 ¡ x2 + 6 = (x ¡ 1)2(2x + a) + bx + c for all x.

2 Find constants a and b if:

a z4 + 4 = (z2 + az + 2)(z2 + bz + 2) for all z

b 2z4 + 5z3 + 4z2 + 7z + 6 = (z2 + az + 2)(2z2 + bz + 3) for all z.

3 Show that z4 +64 can be factorised into two real quadratic factors of the form z2 +az+8 and

z2 + bz + 8, but cannot be factorised into two real quadratic factors of the form z2 + az + 16

and z2 + bz + 4.

4 Find real numbers a and b such that x4 ¡ 4x2 + 8x ¡ 4 = (x2 + ax + 2)(x2 + bx ¡ 2).

Hence solve the equation x4 + 8x = 4x2 + 4.

5 a (2z ¡ 3) is a factor of 2z3 ¡ z2 + az ¡ 3. Find a 2 R and all zeros of the cubic.

b (3z+2) is a factor of 3z3 ¡ z2 + (a+1)z+a. Find a 2 R and all the zeros of the cubic.

6 a (2x + 1) and (x ¡ 2) are factors of P (x) = 2x4 + ax3 + bx2 ¡ 12x ¡ 8.

Find constants a and b, and all zeros of P (x).

b (x + 3) and (2x ¡ 1) are factors of 2x4 + ax3 + bx2 + ax + 3.

Find constants a and b, and hence determine all zeros of the quartic.

7 a x3 + 3x2 ¡ 9x + c, c 2 R , has two identical linear factors. Prove that c is either 5 or ¡27,

and factorise the cubic into linear factors in each case.

b 3x3 + 4x2 ¡ x + m, m 2 R , has two identical linear factors. Find the possible values of m,

and find the zeros of the polynomial in each case.

There are many theorems about polynomials, some of which we look at now. Some of the theorems are

true for all polynomials, while others are true only for real polynomials.

THE REMAINDER THEOREM

Consider the cubic polynomial P (x) = x3 + 5x2 ¡ 11x + 3.

If we divide P (x) by x ¡ 2, we find that

x3 + 5x2 ¡ 11x+ 3

x¡ 2= x2 + 7x + 3 +

9

x¡ 2

So, when P (x) is divided by x ¡ 2, the remainder is 9.

Notice that P (2) = 8 + 20 ¡ 22 + 3

= 9, which is the remainder.

By considering other examples like the one above, we formulate

the Remainder theorem.

POLYNOMIAL THEOREMSE

A real polynomial is a

polynomial with real

coefficients.

remainder


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



The Remainder Theorem

When a polynomial P (x) is divided by x ¡ k until a constant remainder R

is obtained, then R = P (k).

Proof: By the division algorithm, P (x) = Q(x)(x ¡ k) + R

Letting x = k, P (k) = Q(k) £ 0 + R

) P (k) = R

When using the Remainder theorem, it is important to realise that the following statements are equivalent:

² P (x) = (x ¡ k)Q(x) + R

² P (k) = R

² P (x) divided by x ¡ k leaves a remainder of R.

Use the Remainder theorem to find the remainder when x4 ¡ 3x3 + x¡ 4 is divided by x+ 2.

If P (x) = x4 ¡ 3x3 + x ¡ 4, then

P (¡2) = (¡2)4 ¡ 3(¡2)3 + (¡2) ¡ 4

= 16 + 24 ¡ 2 ¡ 4

= 34

) when x4 ¡ 3x3 + x¡ 4 is divided by x+ 2, the remainder is 34. fRemainder theoremg

When P (x) is divided by x2 ¡ 3x + 7, the quotient is x2 + x ¡ 1 and the remainder R(x)

is unknown.

When P (x) is divided by x ¡ 2 the remainder is 29. When P (x) is divided by x + 1 the

remainder is ¡16.

Find R(x) in the form ax + b.

When the divisor is x2 ¡ 3x + 7, P (x) = (x2 + x ¡ 1)| {z }Q(x)

(x2 ¡ 3x + 7)| {z }D(x)

+ ax + b| {z }R(x)

.

Now P (2) = 29 fRemainder theoremg) (22 + 2 ¡ 1)(22 ¡ 6 + 7) + 2a + b = 29

) (5)(5) + 2a + b = 29

) 2a + b = 4 .... (1)

Also, P (¡1) = ¡16 fRemainder theoremg) ((¡1)2 + (¡1) ¡ 1)((¡1)2 ¡ 3(¡1) + 7) + (¡a + b) = ¡16

) (¡1)(11) ¡ a + b = ¡16

) ¡a + b = ¡5 .... (2)

Solving (1) and (2) simultaneously gives a = 3 and b = ¡2.

) R(x) = 3x ¡ 2.




0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



EXERCISE 6E.1

1 For P (x) a real polynomial, write two equivalent statements for each of:

a If P (2) = 7, then ....

b

c If P (x) divided by x ¡ 5 has a remainder of 11 then ....

2 Without performing division, find the remainder when:

a x3 + 2x2 ¡ 7x + 5 is divided by x ¡ 1

b x4 ¡ 2x2 + 3x ¡ 1 is divided by x + 2.

3 Find a 2 R given that:

a when x3 ¡ 2x + a is divided by x ¡ 2, the remainder is 7

b when 2x3 + x2 + ax ¡ 5 is divided by x + 1, the remainder is ¡8.

4 When x3 + 2x2 + ax + b is divided by x ¡ 1 the remainder is 4, and when divided by x + 2the remainder is 16. Find constants a and b.

5 2xn +ax2 ¡6 leaves a remainder of ¡7 when divided by x¡1, and 129 when divided by x+3.

Find a and n given that n 2 Z+.

6 When P (z) is divided by z2 ¡ 3z + 2 the remainder is 4z ¡ 7.

Find the remainder when P (z) is divided by: a z ¡ 1 b z ¡ 2.

7 When P (z) is divided by z + 1 the remainder is ¡8, and when divided by z ¡ 3 the remainder

is 4. Find the remainder when P (z) is divided by (z ¡ 3)(z + 1).

8 If P (x) is divided by (x ¡ a)(x ¡ b), where a 6= b, a, b 2 R , prove that the remainder is:³P (b)¡ P (a)

b¡ a

´£ (x ¡ a) + P (a).

THE FACTOR THEOREM

For any polynomial P (x), k is a zero of P (x) , (x ¡ k) is a factor of P (x).

Proof: k is a zero of P (x) , P (k) = 0 fdefinition of a zerog, R = 0 fRemainder theoremg, P (x) = Q(x)(x ¡ k) fdivision algorithmg, (x ¡ k) is a factor of P (x) fdefinition of a factorg

The Factor theorem says that if 2 is a zero of P (x) then (x¡2) is a factor of P (x), and vice versa.

If P (x) = Q(x)(x + 3) ¡ 8, then ....


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Find k given that (x ¡ 2) is a factor of x3 + kx2 ¡ 3x + 6.

Hence, fully factorise x3 + kx2 ¡ 3x + 6.

Let P (x) = x3 + kx2 ¡ 3x + 6

Since (x ¡ 2) is a factor, P (2) = 0 fFactor theoremg) (2)3 + k(2)2 ¡ 3(2) + 6 = 0

) 8 + 4k = 0

) k = ¡2

We now use either the division algorithm or synthetic division to find the other factors of P (x):

Division algorithm or Synthetic division

Using either method we find that P (x) = (x ¡ 2)(x2 ¡ 3)

= (x ¡ 2)(x +p

3)(x ¡p

3)

EXERCISE 6E.2

1 Find the constant k and hence factorise the polynomial if:

a 2x3 + x2 + kx ¡ 4 has the factor (x + 2)

b x4 ¡ 3x3 ¡ kx2 + 6x has the factor (x ¡ 3).

2 Find constants a and b given that 2x3 + ax2 + bx + 5 has factors (x ¡ 1) and (x + 5).

3 a Suppose 3 is a zero of P (z) = z3 ¡ z2 + (k ¡ 5)z + (k2 ¡ 7).

Find the possible values of k 2 R and all the corresponding zeros of P (z).

b Show that (z ¡ 2) is a factor of P (z) = z3 +mz2 + (3m ¡ 2)z ¡ 10m ¡ 4 for all values

of m 2 R . For what value(s) of m is (z ¡ 2)2 a factor of P (z)?

4 a Consider P (x) = x3 ¡ a3 where a is real.

i Find P (a). What is the significance of this result?

ii Factorise x3 ¡ a3 as the product of a real linear and a quadratic factor.

b Now consider P (x) = x3 + a3, where a is real.

i Find P (¡a). What is the significance of this result?

ii Factorise x3 + a3 as the product of a real linear and a quadratic factor.


x2 + 0x ¡ 3

x ¡ 2 x3 ¡ 2x2 ¡ 3x + 6

¡ (x3 ¡ 2x2)

0x2 ¡ 3x

¡ (0x2 ¡ 0x)

¡3x + 6

¡(¡3x + 6)

0

2 1 ¡2 ¡3 6

0 2 0 ¡6

1 0 ¡3 0


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



5 a Prove that “x + 1 is a factor of xn + 1, n 2 Z , n is odd”.

b Find the real number a such that (x ¡ 1 ¡ a) is a factor of P (x) = x3 ¡ 3ax ¡ 9.

THE FUNDAMENTAL THEOREM OF ALGEBRA

The theorems we have just seen for real polynomials can be generalised

in the Fundamental Theorem of Algebra:

a Every polynomial of degree n > 1 has at least one zero which

can be written in the form a + bi where a, b 2 R .

b If P (x) is a polynomial of degree n, then P (x) has exactly n

zeros, some of which may be either irrational numbers or complex

numbers.

Using the Fundamental Theorem of Algebra, the following properties of real polynomials can be

established:

² Every real polynomial of degree n can be factorised into n complex linear factors, some of which

may be repeated.

² Every real polynomial can be expressed as a product of real linear and real irreducible quadratic

factors (where ¢ < 0).

² Every real polynomial of degree n has exactly n zeros, some of which may be repeated.

² If p + qi (q 6= 0) is a zero of a real polynomial then its complex conjugate p ¡ qi is also a

zero.

² Every real polynomial of odd degree has at least one real zero.

Suppose ¡3 + i is a zero of P (x) = ax3 + 9x2 + ax ¡ 30 where a is real.

Find a and hence find all zeros of the cubic.

Since P (x) is real, both ¡3 + i and ¡3 ¡ i are zeros.

These have sum = ¡6 and product = (¡3 + i)(¡3 ¡ i) = 10.

) the zeros ¡3 § i come from the quadratic x2 + 6x + 10.

The coefficient of x3 is a. The constant term is 10£ (¡3) = ¡30.

Consequently, ax3 + 9x2 + ax ¡ 30 = (x2 + 6x + 10)(ax ¡ 3)

= ax3 + (6a ¡ 3)x2 + (10a ¡ 18)x ¡ 30

Equating coefficients of x2 gives: 6a ¡ 3 = 9 ) a = 2

Equating coefficients of x gives: 10a ¡ 18 = a ) a = 2

) a = 2 and the linear factor is (2x ¡ 3)

) the other two zeros are ¡3 ¡ i and 3

2.


Gauss proved this

theorem in as

his PhD dissertation.

1799


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



One zero of ax3 + (a + 1)x2 + 10x + 15, a 2 R , is purely imaginary.

Find a and the zeros of the polynomial.

Let the purely imaginary zero be bi, b 6= 0.

Since P (x) is real, both bi and ¡bi are zeros.

These have sum = 0 and product = ¡b2i2 = b2

) the zeros §bi come from the quadratic x2 + b2.

The coefficient of x3 is a. The constant term is b2 £ 15

b2= 15.

Consequently, ax3 + (a + 1)x2 + 10x + 15 = (x2 + b2)(ax +15

b2)

= ax3 +³15

b2

´x2 + ab2x + 15

Equating coefficients of x2 gives: a + 1 =15

b2.... (1)

Equating coefficients of x gives: ab2 = 10 .... (2)

) ab2 + b2 = 15 fusing (1)g) 10 + b2 = 15 fusing (2)g

) b2 = 5

) b = §p

5

In (2), since b2 = 5, 5a = 10 ) a = 2.

The linear factor ax +15

b2becomes 2x + 3

) a = 2 and the zeros are §ip

5, ¡3

2.

EXERCISE 6E.3

1 Find all real polynomials of degree 3 with zeros ¡1

2and 1 ¡ 3i.

2 p(x) is a real cubic polynomial for which p(1) = p(2 + i) = 0 and p(0) = ¡20.

Find p(x) in expanded form.

3 2 ¡ 3i is a zero of P (z) = z3 + pz + q where p and q are real.

a Using conjugate pairs, find p and q and the other two zeros.

b Check your answer by solving for p and q using P (2 ¡ 3i) = 0.

4 3 + i is a root of z4 ¡ 2z3 + az2 + bz + 10 = 0, where a and b are real. Find a and b and the

other roots of the equation.

5 One zero of P (z) = z3 + az2 + 3z + 9, a 2 R , is purely imaginary. Find a and hence factorise

P (z) into the product of linear factors.

6 At least one zero of P (x) = 3x3 + kx2 + 15x + 10, k 2 R , is purely imaginary. Find k and

hence factorise P (x) into the product of linear factors.



0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



SUM AND PRODUCT OF ROOTS THEOREM

We have seen that for the quadratic equation ax2 + bx + c = 0, a 6= 0, the sum of the roots is ¡ b

a

and the product of the roots isc

a.

For the polynomial equation anxn + an¡1x

n¡1 + :::: + a2x2 + a1x + a0 = 0, an 6= 0

which can also be written asnP

r=0

arxr = 0,

the sum of the roots is¡an¡1

an, and the product of the roots is

(¡1)na0an

.

We can explain this result as follows:

Consider the polynomial equation an(x ¡ ®1)(x ¡ ®2)(x ¡ ®3)::::(x ¡ ®n) = 0

with roots ®1, ®2, ®3, ...., ®n.

Expanding the LHS we have

an (x ¡ ®1)(x ¡ ®2)| {z }(x ¡ ®3)(x ¡ ®4)::::(x ¡ ®n)

= an (x2 ¡ [®1 + ®2]x + (¡1)2®1®2)(x ¡ ®3)| {z }(x ¡ ®4)::::(x ¡ ®n)

= an(x3 ¡ [®1 + ®2 + ®3]x2 + :::: + (¡1)3®1®2®3)(x ¡ ®4)::::(x ¡ ®n)

fthe term of order x is no longer important as it will not contribute to

either the xn¡1 term or the constant termg...

= an(xn ¡ [®1 + ®2 + ®3 + :::: + ®n]xn¡1 + :::: + (¡1)n®1®2®3:::: ®n)

= anxn ¡ an[®1 + ®2 + ®3 + :::: + ®n]xn¡1 + :::: + (¡1)n®1®2®3:::: ®nan

Equating coefficients,

an¡1 = ¡an[®1 + ®2 + ®3 + :::: + ®n] and a0 = (¡1)n®1®2®3:::: ®nan

) ¡an¡1

an= ®1 + ®2 + ®3 + :::: + ®n and

(¡1)na0

an= ®1®2®3:::: ®n

Find the sum and product of the roots of 2x3 ¡ 7x2 + 8x ¡ 1 = 0.

The sum of the roots = ¡ (¡7)

2= 7

2.

The polynomial equation has degree 3.

) the product of the roots =(¡1)3(¡1)

2= 1

2.



0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



A real polynomial has the form P (x) = 3x4 ¡ 12x3 + cx2 + dx+ e. The graph of y = P (x)

has y-intercept 180. It cuts the x-axis at 2 and 6, and does not meet the x-axis anywhere else.

Suppose the other two zeros are m § ni, n > 0. Use the sum and product formulae to find m

and n.

If the other two zeros are m § ni, the sum of the zeros is

2 + 6 + (m + ni) + (m ¡ ni) = ¡¡12

3

) 2m + 8 = 4

) m = ¡2

The constant term e is the y-intercept.

) e = 180

The product of the zeros is 2 £ 6 £ (¡2 + ni)(¡2 ¡ ni) =(¡1)4180

3= 60

) 12(4 + n2) = 60

) 4 + n2 = 5

) n2 = 1

) n = 1 fas n > 0gSo, m = ¡2 and n = 1.

EXERCISE 6E.4

1 Find the sum and product of the roots of:

a 2x2 ¡ 3x + 4 = 0 b 3x3 ¡ 4x2 + 8x ¡ 5 = 0

c x4 ¡ x3 + 2x2 + 3x ¡ 4 = 0 d 2x5 ¡ 3x4 + x2 ¡ 8 = 0

e x7 ¡ x5 + 2x ¡ 9 = 0 f x6 ¡ 1 = 0

2 A real cubic polynomial P (x) has zeros 3§ ip

2 and 2

3. It has a leading coefficient of 6. Find:

a the sum and product of its zeros b the coefficient of x2

c the constant term.

3 A real polynomial of degree 5 has leading coefficient ¡1 and zeros of ¡2, 3 § i, andpk § 1.

The y-intercept is 18. Find:

a k b the coefficient of x4.

4 A real polynomial of degree 5 has leading coefficient 2 and the coefficient of x4 is 3. When the

polynomial is graphed, the y-intercept is 5, and it cuts the x-axis at 1

2and 1 § p

2 only.

Suppose the other two zeros are m § ni, n > 0. Use the sum and product formulae to find m

and n.

5 A real quartic polynomial has leading coefficient 1 and zeros of the form a§ i and 3 § a, where

a 2 R . Its constant term is 25. What are the possible values that a may take?

6 x3 ¡ px2 + qx ¡ r = 0 has non-zero roots p, q, and r, where p, q, r 2 R .

a Show that q = ¡r and p = ¡1

r. b Hence find p, q, and r.



0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


CUBIC GRAPHSINVESTIGATION 2


Any polynomial with real coefficients can be graphed on the Cartesian plane.

Use of a graphics calculator or the graphing package provided will help in this section.

CUBIC POLYNOMIALS

Every real cubic polynomial can be categorised into one of four types. In each case a 2 R , a 6= 0,

and the zeros are ®, ¯, °.

Type 1: Three real, distinct zeros: P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °)

Type 2: Two real zeros, one repeated: P (x) = a(x ¡ ®)2(x ¡ ¯)

Type 3: One real zero repeated three times: P (x) = a(x ¡ ®)3

Type 4: One real and two complex conjugate zeros:

P (x) = (x ¡ ®)(ax2 + bx + c), ¢ = b2 ¡ 4ac < 0.

What to do:

1 Experiment with the graphs of Type 1 cubics. State the effect of changing both the size and

sign of a. What is the geometrical significance of ®, ¯, and °?

2 Experiment with the graphs of Type 2 cubics. What is the geometrical significance of the squared

factor?

3 Experiment with the graphs of Type 3 cubics. What is the geometrical significance

of ®?

4 Experiment with the graphs of Type 4 cubics. What is the geometrical significance

of ® and the quadratic factor which has complex zeros?

From Investigation 2 you should have discovered that:

GRAPHING REAL POLYNOMIALSF

² If a > 0, the graph has shape or . If a < 0 it is or .

² All cubics are continuous smooth curves.

² Every cubic polynomial must cut the x-axis at least once, and so has at least one real zero.

² For a cubic of the form P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °),

®, ¯, ° 2 R , the graph has three distinct x-intercepts

corresponding to the three distinct zeros ®, ¯, and °. The

graph crosses over or cuts the x-axis at these points, as

shown.

² For a cubic of the form P (x) = a(x ¡ ®)2(x ¡ ¯), ®,

¯ 2 R , the graph touches the x-axis at the repeated zero ®

and cuts it at the other x-intercept ¯, as shown.

GRAPHING

PACKAGE

x

x

® ¯ °

¯ ®


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



² For a cubic of the form P (x) = a(x ¡ ®)3, x 2 R , the

graph has only one x-intercept, ®. The graph is horizontal

at this point, and the x-axis is a tangent to the curve even

though the curve crosses over it.

² For a cubic of the form P (x) = (x ¡ ®)(ax2 + bx + c)

where ¢ < 0, there is only one x-intercept, ®. The graph

cuts the x-axis at this point. The other two zeros are complex

and so do not appear on the graph.

Find the equation of the cubic with graph:

a b

a The x-intercepts are ¡1, 2, 4.

) y = a(x + 1)(x ¡ 2)(x ¡ 4)

But when x = 0, y = ¡8

) a(1)(¡2)(¡4) = ¡8

) a = ¡1

So, y = ¡(x + 1)(x ¡ 2)(x ¡ 4)

b The graph touches the x-axis at 2

3,

indicating a squared factor (3x¡ 2)2.

The other x-intercept is ¡3,

so y = a(3x ¡ 2)2(x + 3).

But when x = 0, y = 6

) a(¡2)2(3) = 6

) a = 1

2

So, y = 1

2(3x ¡ 2)2(x + 3)

When determining a polynomial function from a given graph, if we are not given all the zeros, or if some

of the zeros are complex, we write a factor in general form.

For example: ² If an x-intercept is not given, use

P (x) = (x ¡ k)2 (ax + b)| {z } .

most general form of a linear

Using P (x) = a(x ¡ k)2(x + b) is

more complicated.

² If there is clearly only one x-intercept

and that is given, use

P (x) = (x ¡ k) (ax2 + bx + c)| {z } .

most general form of a quadratic

¢ < 0


x

k?

y

x-1

-8

2 4

y

x-3

6

We

x

k

Either graph is possible.

x

x

®

®


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Find the equation of the cubic which cuts the x-axis at 2 and ¡3, cuts the y-axis at ¡48, and

which passes through the point (1, ¡40).

The zeros are 2 and ¡3, so y = (x ¡ 2)(x + 3)(ax + b), a 6= 0.

When x = 0, y = ¡48

) (¡2)(3)b = ¡48

) b = 8

When x = 1, y = ¡40

) (¡1)(4)(a + 8) = ¡40

) a + 8 = 10

) a = 2

So, the equation is y = (x ¡ 2)(x + 3)(2x + 8)

) y = 2(x ¡ 2)(x + 3)(x + 4)

EXERCISE 6F.1

1 For a cubic polynomial P (x), state the geometrical significance of:

a a single real linear factor such as (x ¡ ®), ® 2 R

b a squared real linear factor such as (x ¡ ®)2, ® 2 R

c a cubed real linear factor such as (x ¡ ®)3, ® 2 R .

2 Find the equation of the cubic with graph:

a b c

d e f

3 Find the equation of the cubic whose graph:

a cuts the x-axis at 3, 1, and ¡2, and passes through (2, ¡4)

b cuts the x-axis at ¡2, 0, and 1

2, and passes through (¡3, ¡21)

c touches the x-axis at 1, cuts the x-axis at ¡2, and passes through (4, 54)

d touches the x-axis at ¡2

3, cuts the x-axis at 4, and passes through (¡1, ¡5).


y

x-1

12

2 3

y

x-3

- Qw Qw6

y

x-4

-12

3

y

x-3 -2

-12

- Qw

y

x-4

9

3

y

x-5

-2

-5

5


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


QUARTIC GRAPHSINVESTIGATION 3


4 Match the given graphs to the corresponding cubic function:

a y = 2(x ¡ 1)(x + 2)(x + 4) b y = ¡(x + 1)(x ¡ 2)(x ¡ 4)

c y = (x ¡ 1)(x ¡ 2)(x + 4) d y = ¡2(x ¡ 1)(x + 2)(x + 4)

e y = ¡(x ¡ 1)(x + 2)(x + 4) f y = 2(x ¡ 1)(x ¡ 2)(x + 4)

A B C

D E F

5 Find the equation of a real cubic polynomial which:

a cuts the x-axis at 1

2and ¡3, cuts the y-axis at 30, and passes through (1, ¡20)

b cuts the x-axis at 1, touches the x-axis at ¡2, and cuts the y-axis at (0, 8)

c cuts the x-axis at 2, cuts the y-axis at ¡4, and passes through (1, ¡1) and (¡1, ¡21).

QUARTIC POLYNOMIALS

There are considerably more possible factor types to consider for quartic functions. We will consider

quartics containing certain types of factors.

What to do:

1 Experiment with the graphs of quartics which have:

a four distinct real linear factors

b a squared real linear factor and two distinct real linear factors

c two squared real linear factors

d a cubed real linear factor and one distinct real linear factor

e a real linear factor raised to the fourth power

f one real quadratic factor with ¢ < 0 and two real linear factors

g two real quadratic factors each with ¢ < 0.

2 Summarise your observations.

y

x

-4

8

1 2x

y

-4

16

1 2

x

y

-1

-8

24

x

y

-4-2

-16

1x

y

-4-2

16

1

x

y

-4-2

8

1

GRAPHING

PACKAGE


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



From Investigation 3 you should have discovered that:

² For a quartic polynomial in which a is the coefficient of x4:

I If a > 0 the graph opens upwards.

I If a < 0 the graph opens downwards.

² If a quartic with a > 0 is fully factorised into real linear factors, then:

I for a single factor (x ¡ ®), the

graph cuts the x-axis at ®

I for a square factor (x ¡ ®)2, the

graph touches the x-axis at ®

I for a cubed factor (x ¡ ®)3, the

graph cuts the x-axis at ® and is

‘flat’ at ®

I for a quadruple factor (x ¡ ®)4,

the graph touches the x-axis and is

‘flat’ at that point.

I If a quartic with a > 0 has one

real quadratic factor with ¢ < 0we could have:

I If a quartic with a > 0 has two real

quadratic factors both with ¢ < 0we have:

The graph does not meet the x-axis

at all.

Find the equation of the quartic

with graph:

The graph touches the x-axis at ¡1and cuts it at ¡3 and 3.

) y = a(x + 1)2(x + 3)(x ¡ 3), a 6= 0

But when x = 0, y = ¡3

) ¡3 = a(1)2(3)(¡3)

) ¡3 = ¡9a

) a = 1

3

) y = 1

3(x + 1)2(x + 3)(x ¡ 3)


x

y®

x

y ®

x

y®

x

y®

x

y

x

y

y

x-3-1

-3

3

x

y ®

or


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Find the quartic which touches the x-axis at 2, cuts the x-axis at ¡3, and also passes

through (1, ¡12) and (3, 6).

The graph touches the x-axis at 2, so (x ¡ 2)2 is a factor.

The graph cuts the x-axis at ¡3, so (x + 3) is a factor.

) P (x) = (x ¡ 2)2(x + 3)(ax + b) where a and b are constants, a 6= 0.

Now P (1) = ¡12

) (¡1)2(4)(a + b) = ¡12

) a + b = ¡3 .... (1)

and P (3) = 6,

) 12(6)(3a + b) = 6

) 3a + b = 1 .... (2)

Solving (1) and (2) simultaneously gives a = 2, b = ¡5.

) P (x) = (x ¡ 2)2(x + 3)(2x ¡ 5)

EXERCISE 6F.2

1 Find the equation of the quartic with graph:

a b c

d e f

2 Match the given graphs to the corresponding quartic functions:

a y = (x ¡ 1)2(x + 1)(x + 3) b y = ¡2(x ¡ 1)2(x + 1)(x + 3)

c y = (x ¡ 1)(x + 1)2(x + 3) d y = (x ¡ 1)(x + 1)2(x ¡ 3)

e y = ¡1

3(x ¡ 1)(x + 1)(x + 3)2 f y = ¡(x ¡ 1)(x + 1)(x ¡ 3)2

A B C


y

x-1 1

2

y

x-2

-1

-16

2

y

x-3 -1

-6

We

y

x-1

-16

4y

x-3-1

-9

3

Ew

y

x-2 3

(-3 54),

y

x

-3-1

1

y

x-3 -11

y

x-3 -1 1


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


GENERAL POLYNOMIALSDISCUSSION

ACTIVITY


D E F

3 Find the equation of the quartic whose graph:

a cuts the x-axis at ¡4 and 1

2, touches it at 2, and passes through the point (1, 5)

b touches the x-axis at 2

3and ¡3, and passes through the point (¡4, 49)

c cuts the x-axis at §1

2and §2, and passes through the point (1, ¡18)

d touches the x-axis at 1, cuts the y-axis at ¡1, and passes through (¡1, ¡4) and (2, 15).

Click on the icon to run a card game for graphs of cubic and quartic functions.

GENERAL POLYNOMIALS

We have already seen that every real cubic polynomial must cut the x-axis at least once, and so has at

least one real zero.

If the exact value of the zero is difficult to find, we can use technology to help us. We can then factorise

the cubic as a linear factor times a quadratic, and if necessary use the quadratic formula to find the other

zeros.

This method is particularly useful if we have one rational zero and two irrational zeros that are radical

conjugates.

Consider the general polynomial P (x) = anxn + an¡1x

n¡1 + :::: + a1x + a0, an 6= 0.

² Discuss the behaviour of the graph as x ! ¡1 and x ! 1 depending on

I the sign of an I whether n is odd or even.

² Under what circumstances is P (x) an:

I odd function I even function?

y

x-3-1

1

y

x

-11

3

y

x-1 1 3

CARD GAME


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Find exactly the zeros of P (x) = 3x3 ¡ 14x2 + 5x + 2.

Using the calculator we search for any rational zero.

In this case, x ¼ 0:666 667 or 0:6 indicates x = 2

3is a zero.

) (3x ¡ 2) is a factor.

) 3x3 ¡ 14x2 + 5x + 2 = (3x ¡ 2)(x2 + ax ¡ 1) for some a

= 3x3 + (3a ¡ 2)x2 + (¡3 ¡ 2a)x + 2

Equating coefficients: 3a ¡ 2 = ¡14 and ¡3 ¡ 2a = 5

) 3a = ¡12 and ¡2a = 8

) a = ¡4

) P (x) = (3x ¡ 2)(x2 ¡ 4x ¡ 1) which has zeros 2

3and 2 § p

5 fquadratic formulag

Find exactly the roots of 6x3 + 13x2 + 20x + 3 = 0.

Using technology, x ¼ ¡0:166 666 67 = ¡1

6is a root, so (6x + 1) is a factor of the cubic.

) (6x + 1)(x2 + ax + 3) = 0 for some constant a.

Equating coefficients of x2: 1 + 6a = 13

) 6a = 12

) a = 2

Equating coefficients of x: a + 18 = 20 X

) (6x + 1)(x2 + 2x + 3) = 0

) x = ¡1

6or ¡1 § i

p2 fquadratic formulag

For a quartic polynomial P (x) we first need to establish if there are any x-intercepts at all. If there

are not then the polynomial must have four complex zeros. If there are x-intercepts then we can try to

identify linear or quadratic factors.

EXERCISE 6F.3

1 Find exactly all zeros of:

a x3 ¡ 3x2 ¡ 3x + 1 b x3 ¡ 3x2 + 4x ¡ 2

c 2x3 ¡ 3x2 ¡ 4x ¡ 35 d 2x3 ¡ x2 + 20x ¡ 10

e 4x4 ¡ 4x3 ¡ 25x2 + x + 6 f x4 ¡ 6x3 + 22x2 ¡ 48x + 40

2 Find exactly the roots of:

a x3 + 2x2 + 3x + 6 = 0 b 2x3 + 3x2 ¡ 3x ¡ 2 = 0

c x3 ¡ 6x2 + 12x ¡ 8 = 0 d 2x3 + 18 = 5x2 + 9x

e x4 ¡ x3 ¡ 9x2 + 11x + 6 = 0 f 2x4 ¡ 13x3 + 27x2 = 13x + 15



GRAPHING

PACKAGE


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\208IB_HL-3ed_06.cdr Wednesday, 9 May 2012 9:19:03 AM BEN

NON-CALCULATORREVIEW SET 6A


3 Factorise into linear factors with exact values:

a x3 ¡ 3x2 + 4x ¡ 2 b x3 + 3x2 + 4x + 12

c 2x3 ¡ 9x2 + 6x ¡ 1 d x3 ¡ 4x2 + 9x ¡ 10

e 4x3 ¡ 8x2 + x + 3 f 3x4 + 4x3 + 5x2 + 12x ¡ 12

g 2x4 ¡ 3x3 + 5x2 + 6x ¡ 4 h 2x3 + 5x2 + 8x + 20

4 The following cubics will not factorise neatly. Find their zeros using technology.

a x3 + 2x2 ¡ 6x ¡ 6 b x3 + x2 ¡ 7x ¡ 8

5 A scientist is trying to design a crash test barrier with the characteristics shown graphically below.

The independent variable t is the time

after impact, measured in milliseconds,

such that 0 6 t 6 700.

The dependent variable is the distance

the barrier is depressed during the

impact, measured in millimetres.

a The equation for this graph has the form f(t) = kt(t ¡ a)2, 0 6 t 6 700.

Use the graph to find a. What does this value represent?

b If the ideal crash barrier is depressed by 85 mm after 100 milliseconds, find the value of k.

Hence find the equation of the graph given.

6 Last year, the volume of water in a particular reservoir could be described by the model

V (t) = ¡t3 + 30t2 ¡ 131t + 250 ML, where t is the time in months.

The dam authority rules that if the volume falls below 100 ML, irrigation is prohibited. During

which months, if any, was irrigation prohibited in the last twelve months? Include in your answer

a neat sketch of any graphs you may have used.

7 A ladder of length 10 metres is leaning against a wall so that it is just

touching a cube of edge length one metre as shown.

What height up the wall does the ladder reach?

1 Find real numbers a and b such that:

a a + ib = 4 b (1 ¡ 2i)(a + bi) = ¡5 ¡ 10i c (a + 2i)(1 + bi) = 17 ¡ 19i

2 If z = 3 + i and w = ¡2 ¡ i, find in simplest form:

a 2z ¡ 3w bz¤

wc z3

10 m

1 m

x m

800600400200

12010080604020

f(t)�

t ( )ms


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


CALCULATORREVIEW SET 6B


3 Find exactly the real and imaginary parts of z if z =3

i+p3

+p

3.

4 Find a complex number z such that 2z ¡ 1 = iz ¡ i. Write your answer in the form

z = a + bi where a, b 2 R .

5 Prove that zw¤ ¡ z¤w is purely imaginary or zero for all complex numbers z and w.

6 Given w =z + 1

z¤ + 1where z = a + bi, a, b 2 R , write w in the form x + yi, x, y 2 R .

Hence determine the conditions under which w is purely imaginary.

7 Expand and simplify: a (3x3 + 2x ¡ 5)(4x ¡ 3) b (2x2 ¡ x + 3)2

8 Carry out the following divisions: ax3

x+ 2b

x3

(x+ 2)(x+ 3)

9 Find the sum and product of the zeros of:

a 3x4 ¡ 4x3 + 3x2 + 8 b 2x6 + 2x4 ¡ x3 + 7x ¡ 10

10 State and prove the Remainder theorem.

11 ¡2 + bi is a solution to z2 + az + (3 + a) = 0. Find constants a and b given that they are

real.

12 P (x) has remainder 2 when divided by x ¡ 3, and remainder ¡13 when divided by x + 2.

Find the remainder when P (x) is divided by x2 ¡ x ¡ 6.

13 Find all quartic polynomials with real, rational coefficients, having 2 ¡ ip

3 andp

2 + 1 as

two of the zeros.

14 If f(x) = x3 ¡ 3x2 ¡ 9x + b has (x ¡ k)2 as a factor, show that there are two possible

values of k. For each of these two values of k, find the corresponding value for b, and hence

solve f(x) = 0.

15 Find all real quartic polynomials with rational coefficients, having 3 ¡ ip

2 and 1 ¡ p2 as

two of the zeros.

16 When P (x) = xn + 3x2 + kx+ 6 is divided by (x+ 1) the remainder is 12. When P (x)

is divided by (x ¡ 1) the remainder is 8. Find k and n given that 34 < n < 38.

17 If ® and ¯ are two of the roots of x3 ¡x+1 = 0, show that ®¯ is a root of x3+x2 ¡1 = 0.

Hint: Let x3 ¡ x + 1 = (x ¡ ®)(x ¡ ¯)(x ¡ °).

1 If z =³

5

2¡ i¡ 3 ¡ 2i

´3, x, y 2 Z , express z in the form z = x + yi.

2 Without using a calculator, find z if z2 = 5 ¡ 12i. Check your answer using a calculator.

3 Prove that if z is a complex number then both z + z¤ and zz¤ are real.

4 If z = 4 + i and w = 3 ¡ 2i find 2w¤ ¡ iz.

5 Find rationals a and b such that2¡ 3i

2a+ bi= 3 + 2i.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


REVIEW SET 6C


6 a + ai is a root of x2 ¡ 6x + b = 0 where a, b 2 R .

Explain why b has two possible values. Find a in each case.

7 Find the remainder when x47 ¡ 3x26 + 5x3 + 11 is divided by x + 1.

8 Factorise 2z3 + z2 + 10z + 5 as a product of linear factors with exact terms.

9 A quartic polynomial P (x) has graph y = P (x) which touches the x-axis at (¡2, 0), cuts

the x-axis at (1, 0), cuts the y-axis at (0, 12), and passes through (2, 80).

Find an expression for P (x) in factored form, and hence sketch the graph of y = P (x).

10 Find all zeros of 2z4 ¡ 5z3 + 13z2 ¡ 4z ¡ 6.

11 Factorise z4 + 2z3 ¡ 2z2 + 8 into linear factors.

12 Find the general form of all real polynomials of least degree which have zeros 2 + i and

¡1 + 3i.

13 3 ¡ 2i is a zero of z4 + kz3 + 32z + 3k ¡ 1, where k is real.

Find k and all zeros of the quartic.

14 Find all the zeros of the polynomial z4 + 2z3 + 6z2 + 8z + 8 given that one of the zeros is

purely imaginary.

15 When a polynomial P (x) is divided by x2 ¡ 3x + 2 the remainder is 2x + 3.

Find the remainder when P (x) is divided by x ¡ 2.

16 Find possible values of k if the line with equation y = 2x+ k does not meet the circle with

equation x2 + y2 + 8x ¡ 4y + 2 = 0.

17 P (x) = 2x4 ¡ 8x3 + ax2 + bx ¡ 110, a, b 2 R , has zeros m § 2i and 1 § np

3 where

m, n 2 R .

a Find m and n. b Sketch the graph of y = P (x).

1 Find real numbers x and y such that (3x + 2yi)(1 ¡ i) = (3y + 1)i ¡ x.

2 Solve the equation: z2 + iz + 10 = 6z.

3 Prove that zw¤ + z¤w is real for all complex numbers z and w.

4 Find real x and y such that:

a x + iy = 0 b (3 ¡ 2i)(x + i) = 17 + yi c (x + iy)2 = x ¡ iy

5 z and w are non-real complex numbers with the property that both z + w and zw are real.

Prove that z¤ = w.

6 Find the sum and product of the roots of:

a 2x3 + 3x2 ¡ 4x + 6 = 0 b 4x4 = x2 + 2x ¡ 6

7 Find z ifpz =

2

3¡ 2i+ 2 + 5i.

8 Find the remainder when 2x17 + 5x10 ¡ 7x3 + 6 is divided by x ¡ 2.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



9 5 ¡ i is a zero of 2z3 + az2 + 62z + (a ¡ 5), where a is real. Find a and the other two

zeros.

10 Find, in general form, all real polynomials of least degree which have zeros:

a ip

2 and 1

2b 1 ¡ i and ¡3 ¡ i.

11 P (x) = 2x3 + 7x2 + kx ¡ k is the product of 3 linear factors, 2 of which are identical.

a Show that k can take 3 distinct values.

b Write P (x) as the product of linear factors for the case where k is greatest.

12 Find all roots of 2z4 ¡ 3z3 + 2z2 = 6z + 4.

13 Suppose a and k are real. For what values of k does z3 + az2 + kz + ka = 0 have:

a one real root b 3 real roots?

14 (3x + 2) and (x ¡ 2) are factors of 6x3 + ax2 ¡ 4ax + b. Find a and b.

15 Find the exact values of k for which the line y = x¡k is a tangent to the circle with equation

(x ¡ 2)2 + (y + 3)2 = 4.

16 Find the quotient and remainder when x4 + 3x3 ¡ 7x2 + 11x ¡ 1 is divided by x2 + 2.

Hence, find constants a and b for which x4 + 3x3 ¡ 7x2 + (2 + a)x + b is exactly divisible

by x2 + 2.

17 P (x) is a real polynomial of degree 5 with leading coefficient 1 and zeros m § 2i, 1 § mi,

and 2 (m 2 R ). The graph of y = P (x) cuts the y-axis at ¡56.

a Show that m = §p3. b Find the coefficient of x4.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


Complexnumbers and polynomials - Haese Mathematics · PDF file... we can add, subtract,...

Documents

Transcript of Complexnumbers and polynomials - Haese Mathematics · PDF file... we can add, subtract,...