HW: Pg. 341-342 #13-61 eoo

Quiz 1 Pg. 344 #13-26

Vocabulary• Polynomial Long Division:

– When you divide a polynomial ______ by a divisor _____, you get a quotient polynomial ______ and a remainder polynomial _____.

– This can be written as: • Remainder Theorem:

– If a polynomial _____ is divided by ______, then the remainder is __________.

• Synthetic Division:– Only use the ___________ of the polynomial and the

_____________ must be in the form _________.• Factor Theorem:

– A polynomial _____ has a factor ________ if and only if ___________.

EXAMPLE 1Use polynomial long division

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.

4 3 2Di ( ) 3 5vi x 3de b 4 56 yf x x x x x

EXAMPLE 1Use polynomial long division

Multiply divisor by 3x4/x2 = 3x23x4 – 9x3 + 15x2

4x3 – 15x2 + 4xSubtract.

Bring down next term.

Multiply divisor by 4x3/x2 = 4x4x3 – 12x2 + 20x

–3x2 – 16x – 6Subtract.

Bring down next term.

Multiply divisor by – 3x2/x2 = – 3–3x2 + 9x – 15

–25x + 9

remainder

3x2 + 4x – 3

x2 – 3x + 5 3x4 – 5x3 + 0x2 + 4x – 6) quotient

EXAMPLE 1Use polynomial long division

You can check the result of a division problem by multiplying the quotient by the divisor and adding

the remainder. The result should be the dividend.

(3x2 + 4x – 3)(x2 – 3x + 5) + (–25x + 9)

= 3x2(x2 – 3x + 5) + 4x(x2 – 3x + 5) – 3(x2 – 3x + 5) – 25x + 9

CHECK

= 3x4 – 9x3 + 15x2 + 4x3 – 12x2 + 20x – 3x2 + 9x – 15 – 25x + 9

= 3x4 – 5x3 + 4x – 6

3x4 – 5x3 + 4x – 6

x2 – 3x + 5

= 3x2 + 4x – 3 +–25x + 9

x2 – 3x + 5ANSWER

EXAMPLE 2Use polynomial long division with a linear divisor

x2

x – 2 x3 + 5x2 – 7x + 2)

quotient

x3 – 2x2 Multiply divisor by x3/x = x2.

7x2 – 7x Subtract.

Multiply divisor by 7x2/x = 7x.7x2 – 14x

7x + 2 Subtract.

Multiply divisor by 7x/x = 7.

16remainder

7x – 14

ANSWER

x3 + 5x2 – 7x +2

x – 2= x2 + 7x + 7 +

16

x – 2

3 2( ) 5Divide by x7 2 2f x x x x

+ 7x + 7

GUIDED PRACTICEfor Examples 1 and 2

Divide using polynomial long division.

(2x2 – 3x + 8) + –18x + 7

x2 + 2x – 1

ANSWER

(x2 – 3x + 10) + –30

x + 2ANSWER

4 3 21. 2 12 1 x xx x x

3 22. 24 x 10x x x

EXAMPLE 3Use synthetic division

–3 2 1 –8 5

–6 15 –21

2 –5 7 –16

2x3 + x2 – 8x + 5

x + 3

= 2x2 – 5x + 7 –16

x + 3

ANSWER

SOLUTION

3 2Divide by using sy( ) 2 8 5 nthetx 3 ic division.f x x x x

EXAMPLE 4Factor a polynomial

SOLUTION

Because x + 2 is a factor of f (x), you know that f (–2) = 0. Use synthetic division to find the other

factors.

–2 3 –4 –28 –16

–6 20 16

3 –10 –8 0

3 2Factor completely given that is a fa( ) 3 4 x cto28 6 r2 .1f x x x x

EXAMPLE 4Factor 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 PRACTICEfor Examples 3 and 4

Divide using synthetic division.

x2 + x – 4 +11

x + 3ANSWER

4x2 + 5x + 2 +9

x – 1

ANSWER

3 2 (x 33. ( 4 1) ) x x x

3 2 (x-14 (4 3 7. ) )x x x

3 2( ) 6 5 1 25. f x x x x

3 2( ) 22 406. f x x x x

(x – 4)(x –3)(x + 1)

(x – 4)(x –2)(x +5)

ANSWER

ANSWER

Factor the polynomial completely given that

x – 4 is a factor.

EXAMPLE 5Standardized 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

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.

The correct answer is A. ANSWER

= (x – 3)(x + 5)(x – 4)

= (x – 3)(x2 + x – 20)

GUIDED PRACTICEfor Example 5

Find the other zeros of f given that f (–2) = 0.

3 and –3 1 and –7

ANSWER ANSWER

3 2( ) 2 9 1 87. f x x x x 3 2( ) 8 5 1 48. f x x x x

HOMEWORK:

Pg. 356 #15-35 eoo