1. Use synthetic substitution to evaluate f (x) = x 3 + x 2 – 3x – 10 when x = 2. ANSWER...

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Transcript of 1. Use synthetic substitution to evaluate f (x) = x 3 + x 2 – 3x – 10 when x = 2. ANSWER...

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)