Dividing Polynomials

Warm Up

• Without a calculator, divide the following

113277323

Solution: 49251

This long division technique can also be used to divide polynomials

4

+ 2

2 3 1 2 xxx

Example: Divide x2 + 3x – 2 by x – 1 and check the answer.

x

x2 + x

2x – 2

2x + 2

– 4

remainder

Check:

xx

xxx

22 1.

xxxx 2)1(2.

xxxxx 2)()3( 22 3.

22

2 x

xxx4.

22)1(2 xx5.

4)22()22( xx6.

correct(x + 2)

quotient

(x + 1)

divisor

+ (– 4)

remainder

= x2 + 3x – 2

dividend

Answer: x + 2 +

1x

– 4

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

• (2x² -19x + 8) / (x-8)

2x² - 19x + 8x - 8

Let’s Try One

7

Example: Divide 4x + 2x3 – 1 by 2x – 2 and check the answer.

1 4 0 2 2 2 23 xxxxWrite the terms of the dividend in descending order.

23

2

2x

x

x1.

x2

232 22)22( xxxx 2.

2x3 – 2x2

2233 2)22(2 xxxx 3.

2x2 + 4x

xx

x

2

2 2

4.

+ x

xxxx 22)22( 2 5.

2x2 – 2x

xxxxx 6)22()42( 22 6.

6x – 1

32

6x

x7.

+ 3

66)22(3 xx8.

6x – 6

remainder5)66()16( xx9.

5

Check: (x2 + x + 3)(2x – 2) + 5 = 4x + 2x3 – 1

Answer: x2 + x + 3

22 x

5

Since there is no x2 term in the dividend, add 0x2 as a placeholder.

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8

6 5 2 2 xxx

x

x2 – 2x

– 3x + 6

– 3

– 3x + 6

0

Answer: x – 3 with no remainder.

Check: (x – 2)(x – 3) = x2 – 5x + 6

Example: Divide x2 – 5x + 6 by x – 2.

A Couple of Notes

• 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