6.5 Warm Up

1. Factor 8x3 + 125.

2. Factor 5x3 + 10x2 – x – 2.

3. Factor 200x6 – 2x4.

4. Find the product of (2x – 3)(2x – 5).

5. Find the product of (5x + 2y)3

6.5 Polynomial Division

Long division( ) ( )

( )( ) ( )

n x r xq x

d x d x

Numerator

Denominator Quotient Remainder expressed as a fraction.

Long Division Example( ) ( )

( )( ) ( )

n x r xq x

d x d x

4 3 3 2Divide (2 2 10 9) by ( 5)x x x x x

3 2 4 3

4

3

2

5 2 2 10 9

22 - (2

x

x x x x x

xx x

x

4 32 10 )

---------------------

0 9

x x

Remainder

3 2

9

5x x

Long Division Example 2 3 2 2Divide using long division. 2 2 3 1x x x x

Notice x2 –1 is missing a term.

2 3 2

3

2

0 1 2 2 3x x x x x

xx

x

2 3 2

3

2

0 1 2 2 3

x

x x x x x

xx

x

3 2

2

2

2

-( 0 )

_____________

2 3

22

x x x

x x

x

x

2 3 2

3

2

2

0 1 2 2 3

x

x x x x x

xx

x

3 2

2

2

2

-( 0 )

_____________

2 3 3

22

x x x

x x

x

x

2 - (2 0 2)

____________

x x

3 5x Remainder

2

3 5

1

x

x

Classwork

Textbook page 356 probs 4 – 7.

Synthetic Division

Synthetic division can be used to divide polynomials by an expression in the form of x - k.

Example: Divide (x3 – 8x + 3) by (x + 3).

x + 3 is in the form of x – k.

x + 3 = 0

x = -3 -3 1 0 -8 3

1

-3

-3

9

1

-3

0 Remainder

x2 - 3x + 1Quotient

Classwork

Textbook page 356 probs 8 - 11.

Remainder and Factor Theorem

Remainder Theorem

If a polynomial f(x) is divided by x – k, then the remainder is r = f(k).

Factor Theorem

A polynomial f(x) has a factor x – k if and only if f(k) = 0.

Example: Problem 7 page 116 had a remainder of –7. 2

2

If we evaluate ( ) +2 15 when 4

( 4) ( 4) 2( 4) 15 7

f x x x x

f

f(-4) = the remainder.

Synthetic Division ContinuedExample: Factor (x3 – 8x + 3) given (x + 3) is a factor.

x2 - 3x + 1

1 0 -8 3

1

-3

-3

9

1

-3

0

-3

Continue factoring

( 3) 9 4(1)(1) 3 5

2(1) 2x

So, 3 3 5 3 5

8 3 ( 3)( )( )2 2

x x x x x

Synthetic Division Example 2Given one zero of the polynomial function, find the other zeros.

3 2( ) 2 14 56 40, 10f x x x x x

10 2 -14 –56 -40

220

660

4400

2x2 + 6x + 4

(2x + 4)(x + 1)

(2x + 4) = 0 (x + 1) = 0

x = -2 x = -1

So, the zeros of the polynomial are 10, -1, and –2.

A Geometric Interpretation

3 2( ) 2 14 56 40, 10f x x x x x

-18-15-12 -9 -6 -3 3 6 9 12 15 18

-3920-3430-2940-2450-1960-1470-980-490

490980

-4 -2 2 4 6 8 10 12 14

-10

-8

-6

-4

-2

2

4

6

8

10

The real zeros are the x-intercepts of the graph.

A Geometry ProblemGiven the expression for the volume of a rectangular prism, find an expression for the missing dimension.

x + 5

x + 1

?

3 23 8 45 50V x x x

x = -1 -1 3 8 -45 -50

3-3 -55 -50

500

x = -5 -5 3 5 -50

3-15-10

500

3x -10

Classwork

Textbook page 356 probs 12 - 13.

Homework Textbook page 356 probs 15 – 45 1st col.