6.8 Synthetic Division6.8 Synthetic Division

Polynomial Division, Factors, and Remainders

In this section, we will look at two methods to divide polynomials:

long division (similar to arithmetic long division)

synthetic division (a quicker, short-hand method)

Example: Divide (2x2 + 3x – 4) ÷ (x – 2)

(x – 2) 2x2 + 3x – 4

Rewrite in longdivision form...

divisor

dividendThink, how many timesdoes x go into 2x2 ?

2x

Multiply by the divisor. 2x2 – 4x

Subtract. 7x – 4Think, how many timesdoes x go into 7x ?

+ 7

7x – 14

10 remainder

Write the result like this...

102 7

2x

x

Example: Divide (p3 – 6) ÷ (p – 1)

(p – 1) p3 + 0p2 + 0p – 6

Be sure to add “place-holders”for missing terms...

p2

p3 – p2

p2 + 0p

+ p

p2 – p

p – 6

+ 1

p – 1–5

2 51

1p p

p

Synthetic division can be used when the divisor is in the form (x – k).

Example: Use synthetic division for the following: (2x3– 7x2– 8x + 16) ÷ (x – 4)

First, write down the coefficients in descending order, and k of the divisor in the form x – k :

4 2 –7 –8 16k

2

Bring downthe firstcoefficient.

8

Multiply thisby k

1

Add the column.

4

–4

-16

0These are the coefficients of thequotient (and the remainder)

Repeat the process. 22 4x x

Example: Divide (5x3 + x2 – 7) ÷ (x + 1)

–1 5 1 0 –7Notice thatk is –1 sincesynthetic divisionworks for divisorsin the form (x – k).

place-holder

5

–5

–4

4

4

–4

–11

2 115 4 4

1x x

x

2 2 1 –3 0 –5

f(2) = 23

2

4

5

10

7

14

14

Now, let f(x) = 2x4 + x3 – 3x2 – 5

28

23

What is f(2)?

f(2) = 2(2)4 + (2)3 – 3(2)2 – 5

f(2) = 2(16) + 8 – 3(4) – 5

f(2) = 32 + 8 – 12 – 5

f(2) = 23 This is the same as the remainder when f(x) is divided by (x – 2):

4 1 –6 8 5 13

f(4) = 33

1

4

–2

–8

0

0

5

Example: Use synthetic substitution to find f(4) if f(x) = x4 – 6x3 + 8x2 + 5x + 13

20

33

–2 1 –13 24 108

This means that you can write x3 – 13x2 + 24x + 108 = (x + 2)(x2 – 15x + 54)

1

–2

–15

30

54

–108

0

You can also use synthetic division to find factors of a polynomial...

Example: Given that (x + 2) is a factor of P(x), factor the polynomial P(x) = x3 – 13x2 + 24x + 108

We can use synthetic division to find the other factors...

Factor this... = (x + 2)(x – 9)(x – 6)The complete factorization is: (x + 2)(x – 9)(x – 6)

Since P(–2) = 0, then (x+2) is a factor of P(x)