6-4
Factoring Polynomials
You can use synthetic division to evaluate polynomials.
This process is called synthetic substitution.
The remainder is the value of the polynomial for that x
value.
P(x) = 2x3 + 5x2 – x + 7. Determine P(2).
2
2
2
5 –1
7
4
18 34
9
17 41
Write the coefficients of
the dividend. Use a = 2.
P(2) = 41
Check Substitute 2 for x in P(x) = 2x3 + 5x2 – x + 7.
P(2) = 2(2)3 + 5(2)2 – (2) + 7
P(2) = 41 
Holt Algebra 2
6-4
Factoring Polynomials
Use synthetic substitution to evaluate the
polynomial for the given value.
P(x) = x3 + 3x2 + 4 for x = –3.
–3
1
3 0 4
–3 0 0
1 0 0 4
P(–3) = 4
Write the coefficients of
the dividend. Use 0 for
the coefficient of x2 Use a
= –3.
Check Substitute –3 for x in P(x) = x3 + 3x2 + 4.
P(–3) = (–3)3 + 3(–3)2 + 4
P(–3) = 4 
Holt Algebra 2
6-4
Factoring Polynomials
Synthetic division can also be used to factor a
polynomial.
A binomial (x – a) will be a factor of a
polynomial if and only if the remainder is
zero.
Holt Algebra 2
6-4
Factoring Polynomials
Determine whether the given binomial is a factor
of the polynomial P(x).
A. (x + 1); (x2 – 3x + 1)
Find P(–1) by synthetic
substitution.
–1
1 –3 1
1
–1
–4
4
5
P(–1) = 5
P(–1) ≠ 0, so (x + 1)
is not a factor of
P(x) = x2 – 3x + 1.
Holt Algebra 2
B. (x + 2);
(3x4 + 6x3 – 5x – 10)
Find P(–2) by synthetic
substitution.
–2 3
6
0 –5 –10
–6 0
3
0
0 0 –5
10
0
P(–2) = 0, so (x + 2)
is a factor of P(x) and
P(x) = (x + 2)(3x3 – 5)
6-4
Factoring Polynomials
Determine whether the given binomial is a factor
of the polynomial P(x).
a. (3x – 6);
(3x4 – 6x3 + 6x2 + 3x – 30)
Divide the polynomial by 3,
then find P(2) by synthetic
substitution.
2
1 –2 2 1 –10
1
2 0 4
10
0 2 5
0
P(2) = 0, so (3x – 6) is a factor of P(x)
and P(x) = (3x – 6)(x3 + 2x + 5)
Holt Algebra 2
6-4
Factoring Polynomials
You are already familiar with methods for
factoring quadratic expressions. You can factor
polynomials of higher degrees using many of the
same methods you learned in Lesson 5-3.
Holt Algebra 2
6-4
Factoring Polynomials
Example 2: Factoring by Grouping
Factor: x3 – x2 – 25x + 25.
(x3 – x2) + (–25x + 25)
x2(x – 1) – 25(x – 1)
Group terms.
Factor common monomials
from each group.
(x – 1)(x2 – 25)
Factor out the common
binomial (x – 1).
(x – 1)(x – 5)(x + 5)
Factor the difference of
squares.
Holt Algebra 2
6-4
Factoring Polynomials
Check It Out! Example 2b
Factor: 2x3 + x2 + 8x + 4.
(2x3 + x2) + (8x + 4)
x2(2x + 1) + 4(2x + 1)
Group terms.
Factor common monomials
from each group.
(2x + 1)(x2 + 4)
Factor out the common
binomial (2x + 1).
(2x + 1)(x2 + 4)
Holt Algebra 2
6-4
Factoring Polynomials
HW pg. 433
#’s 17 – 22, 41, 43
Holt Algebra 2