6.8 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)
Think, how many times
does x go into 2x2 ?
divisor
(x – 2)
Rewrite in long
division form...
2x + 7
2x2 + 3x – 4
dividend
2x2 – 4x
Multiply by the divisor.
7x – 4
7x – 14
10
Subtract.
Think, how many times
does x go into 7x ?
Write the result like this...
10
2x  7 
x2
remainder
Example: Divide (p3 – 6) ÷ (p – 1)
2
p + p +1
3
2
(p – 1) p + 0p + 0p – 6
3
Be sure to add “place-holders”
for missing terms...
p – p2
p2 + 0p
p2 – p
p–6
p–1
–5
5
p  p 1
p 1
2
Synthetic division can be used when the divisor is in the
form (x – k).
Example: Use synthetic division for the following:
3
2
(2x – 7x – 8x + 16) ÷ (x – 4)
First, write down the coefficients in descending order, and k of
the divisor in the form x – k :
k
4
Bring down
the first
coefficient.
Multiply this
by k
Add the column.
Repeat the process.
2 –7 –8 16
8
4 -16
2
1
–4
0
These are the coefficients of the
quotient (and the remainder)
2x  x  4
2
Example: Divide (5x3 + x2 – 7) ÷ (x + 1)
Notice that
k is –1 since
synthetic division
works for divisors
in the form (x – k).
place-holder
–1
5
1
–5
5 –4
0 –7
4 –4
4 –11
11
5x  4 x  4 
x 1
2
4
3
2
Now, let f(x) = 2x + x – 3x – 5
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):
2
2
1 –3 0 –5
4 10 14 28
2
5
7 14
23
f(2) = 23
Example: Use synthetic substitution to find f(4) if
4
3
2
f(x) = x – 6x + 8x + 5x + 13
4
1 –6 8
4 –8
1 –2 0
5 13
0 20
5 33
f(4) = 33
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...
–2
1 –13 24 108
–2 30 –108
1 –15 54
0
Since P(–2) = 0, then (x+2)
is a factor of P(x)
This means that you can write
x3 – 13x2 + 24x + 108 = (x + 2)(x2 – 15x + 54)
Factor this...
= (x + 2)(x – 9)(x – 6)
The complete factorization is: (x + 2)(x – 9)(x – 6)