6.3 Dividing Polynomials

Warm Up

• Without a calculator, divide the following

113277323

Solution: 49251

This long division technique can also be used to divide polynomials

POLYNOMIALS – DIVIDINGEX – Long division

• (5x³ -13x² +10x -8) / (x-2)

5x³ - 13x² + 10x - 8x - 2

5x²

5x³ - 10x²- ( )

-3x² + 10x

- 3x

-3x² + 6x- ( )

4x - 8

4x - 8- ( )

+ 4

0

R 0

So in other words…

5x³ -13x² +10x - 8

x-2= 5x² -3x + 4

(5x² -3x + 4) (x-2) = 5x³ -13x² +10x - 8

OR

POLYNOMIALS – DIVIDINGEX#2 – Long division

• (x² +3x -12) / (x+6)

x² + 3x - 12x - 3

x

x² - 3x- ( )

6x -12

+ 6

6x - 18- ( )

6

R 6

• (2x² -19x + 8) / (x-8)

2x² - 19x + 8x - 8

Let’s Try One

• (2x² -19x + 8) / (x-8)

2x² - 19x + 8x - 8

Let’s Try One

EX – Synthetic division(5x³ -13x² +10x -8) / (x-2)

2 5 -13 10 -8

5

10

-3

-6

4

8

0

5x² -3x + 4

Oppositeof number indivisor

EX – Synthetic division(3x³ -4x² +2x -1) / (x+1)

-1 3 -4 2 -1

3

-3

-7

+7

9

-9

-10

3x2 -7x + 9 R-10

Oppositeof number indivisor

Let’s Try One(x³ -13x +12) / (x+4)

EX – Synthetic division(x³ -13x +12) / (x+4)Opposite

of number indivisor

A Couple of Notes

• Use synthetic division when the coefficient in front of x is 1(x- 2) (2x-3)122 xx 122 xx

YES NO

• To test so see if a binomial is a factor, you want to see if you get a remainder of zero. If yes, it is a factor. If you get a remainder, the answer is no.

• (2x² -19x + 8) / (x-8)

2x² - 19x + 8x - 8

From this example, x-8 IS a factor because the remainder is zero

In this case, x-3 is not a factor because there was a remainder of 6

x² + 3x - 12x - 3

x

x² - 3x- ( )

6x -12

+ 6

6x - 18- ( )

6

R 6