Post on 02-Jan-2016
1. Use synthetic substitution to evaluate f (x) = x3 + x2 – 3x – 10 when x = 2.
ANSWER –4
2 1 1 –3 -10
1
2
3
6
3
6
-4
Check HW 5.4multiples of 3
EXAMPLE 1 Use polynomial long division
Divide f (x) = 3x4 – 5x3 + 4x – 6 by x2 – 3x + 5.
SOLUTION
Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coefficient of x2 in the dividend. At each stage, divide the term with the highest power in what is left of the dividend by the first term of the divisor. This gives the next term of the quotient.
EXAMPLE 1 Use polynomial long division
-(3x4 – 9x3 + 15x2)
4x3 – 15x2 + 4x
-(4x3 – 12x2 + 20x)
–3x2 – 16x – 6
-(–3x2 + 9x – 15)
–25x + 9 remainder
3x2 x2 – 3x + 5 3x4 – 5x3 + 0x2 + 4x – 6)
quotient– 3+ 4x
3x4 – 5x3 + 4x – 6x2 – 3x + 5
= 3x2 + 4x – 3 + –25x + 9x2 – 3x + 5
ANSWER
EXAMPLE 2 Use polynomial long division with a linear divisor
Divide f(x) = x3 + 5x2 – 7x + 2 by x – 2.
x2 x – 2 x3 + 5x2 – 7x + 2)
quotient
-(x3 – 2x2)7x2 – 7x
-(7x2 – 14x)
7x + 2
16 remainder
-(7x – 14)
ANSWER x3 + 5x2 – 7x +2x – 2
= x2 + 7x + 7 + 16x – 2
+ 7x + 7
GUIDED PRACTICE for Examples 1 and 2
Divide using polynomial long division.
1. (2x4 + x3 + x – 1) (x2 + 2x – 1)
(2x2 – 3x + 8) + –18x + 7x2 + 2x – 1
ANSWER
(x2 – 3x + 10) + –30 x + 2
ANSWER
2. (x3 – x2 + 4x – 10) (x + 2)
GUIDED PRACTICE for Examples 1 and 2
Divide using polynomial long division.
(x2 – 3x + 10) + –30 x + 2
ANSWER
2. (x3 – x2 + 4x – 10) (x + 2)
1042 23 xxxx
2x
)2( 23 xx xx 43 2
x3
)63( 2 xx
1010 x
10
)2010( x30
EXAMPLE 3Use synthetic division
Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic division.
–3 2 1 –8 5
SOLUTION
2
-6
-5
15
7
-21
-16
3162 752 xxx
EXAMPLE 4 Factor a polynomial
Factor f (x) = 3x3 – 4x2 – 28x – 16 completely given that x + 2 is a factor.
SOLUTION
–2 3 -4 -28 -16
3
-6
-10
20
-8
16
0
8103 2 xx
EXAMPLE 4 Factor a polynomial
Use the result to write f (x) as a product of two factors and then factor completely.
f (x) = 3x3 – 4x2 – 28x – 16 Write original polynomial.
= (x + 2)(3x2 – 10x – 8) Write as a product of two factors.
= (x + 2)(3x + 2)(x – 4) Factor trinomial.
GUIDED PRACTICE for Examples 3 and 4
Divide using synthetic division.
3. (x3 + 4x2 – x – 1) (x + 3)
x2 + x – 4 +11
x + 3ANSWER
–3 1 4 -1 -1
1
-3
1
-3
-4
12
11
GUIDED PRACTICE for Examples 3 and 4
Factor the polynomial completely given that x – 4 is a factor.
5. f (x) = x3 – 6x2 + 5x + 12
4 1 -6 5 12
1
4
-2
-8
-3
-12
0
322 xx
)32)(4( 2 xxx
(x – 4)(x –3)(x + 1)
GUIDED PRACTICE for Examples 5 and 6
Find the other zeros of f given that f (–2) = 0.
7. f (x) = x3 + 2x2 – 9x – 18
–2 1 2 -9 -18
1
-2
0
0
-9
18
0
92 x
0)9)(2( 2 xx
0)3)(3)(2( xxx
Zeros are -2, 3, -3
EXAMPLE 5 Standardized Test Practice
SOLUTION
Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic division.
3 1 –2 –23 60
3 3 –60
1 1 –20 0
EXAMPLE 5
Use the result to write f (x) as a product of two factors. Then factor completely.
f (x) = x3 – 2x2 – 23x + 60
The zeros are 3, –5, and 4.
Standardized Test Practice
The correct answer is A. ANSWER
= (x – 3)(x + 5)(x – 4)
= (x – 3)(x2 + x – 20)
Class/Homework AssignmentWS 5.5 (1-24 mult. of 3)