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Chapter 5
Polynomials and
Factoring
Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
5.2
Factoring Trinomials of the Type
x2 + bx + c
• When the Constant Term Is Positive
• When the Constant Term Is Negative
• Prime Polynomials
• Factoring Completely
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5-2
To Factor x2 + bx + c when c is
Positive
When the constant term of a trinomial
is positive, look for two numbers with
the same sign. The sign is that of the
middle term:
x2 – 7x + 10 (x – 2)(x – 5);
x2 + 7x + 10 (x + 2)(x + 5);
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Example
Factor: x2 + 7x + 12
Solution Think of FOIL in
reverse.
(x + )(x +
)
We need a constant term
that has a product of 12
and a sum of 7. We list
some pairs of numbers
that multiply to 12.
Pairs of
Factors of 12
Sums of
Factors
1, 12
13
2, 6
3, 4
8
7
1, 12
2, 6
3, 4
13
8
7
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5-4
Example
Factor: x2 + 7x + 12
Since 3  4 = 12 and 3 + 4 = 7,
the factorization of x2 + 7x + 12 is
(x + 3)(x + 4).
To check we simply multiply the two binomials.
Check: (x + 3)(x + 4) = x2 + 4x + 3x + 12
= x2 + 7x + 12
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5-5
Example
Factor: y2  8y + 15
Solution
Since the constant term is positive and the coefficient of the
middle term is negative, we look for the factorization of 15 in
which both factors are negative. Their sum must be 8.
Pairs of
Factors of 15
Sums of
Factors
1, 15
3, 5
16
8
Sum of 8
y2  8y + 15 = (y  3)(y  5)
Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
5-6
To Factor x2 + bx + c When c is
Negative
When the constant term of a trinomial is
negative, look for two numbers whose product
is negative. One must be positive and the other
negative:
x2 – 4x – 21 = (x + 3)(x – 7);
x2 + 4x – 21 = (x – 3)(x + 7).
Select the two numbers so that the number with the larger
absolute value has the same sign as b, the coefficient of the
middle term.
Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
Example
Factor: x2  5x  24
Solution The constant term
must be expressed as the
product of a negative
number and a positive
number. Since the sum of
the two numbers must be
negative, the negative
number must have the
greater absolute value.
Pairs of
Factors of 24
Sums of
Factors
1, 24
2, 12
23
10
3, 8
4, 6
6, 4
8, 3
5
2
2
5
x2  5x  24 = (x + 3)(x  8)
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5-8
Example
Factor: t2  32 + 4t
Solution Rewrite the
trinomial t2 + 4t  32. We
need one positive and one
negative factor. The sum
must be 4, so the positive
factor must have the larger
absolute value.
Pairs of
Factors of
32
Sums of
Factors
1, 32
2, 16
4, 8
31
14
4
t2 + 4t  32 = (t + 8)(t  4)
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5-9
Example
Factor: a2 + ab  30b2
Solution
We need the factors of 30b2 that when added
equal b.
Those factors are 5b and 6b.
a2 + ab  30b2 = (a  5b)(a + 6b)
Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
5-10
Prime Polynomials
A polynomial that cannot be factored is considered
prime.
Example: x2  x + 7
Often factoring requires two or more steps.
Remember, when told to factor, we should factor
completely. This means the final factorization
should contain only prime polynomials.
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5-11
Example
Factor: 2x3  24x2 + 72x
Solution
Always look first for a common factor. We can
factor out 2x:
2x(x2  12x + 36)
Since the constant term is positive and the
coefficient of the middle term is negative, we look
for the factorization of 36 in which both factors
are negative. Their sum must be 12.
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5-12
Example
The factorization of
(x2  12x + 36) is
(x  6)(x  6) or (x  6)2
Pairs of
Factors of 36
Sums of
Factors
1, 36
2, 18
37
20
3, 12
4, 9
6, 6
15
13
12
The factorization of
2x3  24x2 + 72x is
2x(x  6)2 or 2x(x  6)(x  6)
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5-13
Once common factors have been factored out,
the following summary can be used.
To Factor x2 + bx + c
1. Find a pair of factors that have c as their
product and b as their sum.
a)If c is positive, its factors will have the same
sign as b.
b)If c is negative, one factor will be positive and
the other will be negative. Select the factors
such that the factor with the larger absolute
value has
the same sign as b.
2. Check by multiplying.
Copyright © 2014, 2010, and 2006 Pearson Education, Inc.