Factorization using Perfect
Square Trinomials
Andrew Gloag
Eve Rawley
Anne Gloag
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Printed: November 6, 2014
AUTHORS
Andrew Gloag
Eve Rawley
Anne Gloag
www.ck12.org
C HAPTER
Chapter 1. Factorization using Perfect Square Trinomials
1
Factorization using Perfect
Square Trinomials
Here you’ll learn how to factor polynomials that are perfect square trinomials. You’ll also learn how to solve
quadratic polynomial equations by factoring.
What if you had a trinomial expression like x2 + 10x + 25 in which the first and third terms were perfect squares and
the second term was twice the product of the square roots of the first and third terms? How could you factor that
expression? After completing this Concept, you’ll be able to factor perfect square trinomials like this one.
Watch This
MEDIA
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URL: http://www.ck12.org/flx/render/embeddedobject/133019
CK-12 Foundation: 0911S Factoring Perfect Square Trinomials
For more examples of factoring perfect square trinomials, watch the videos at http://www.onlinemathlearning.co
m/perfect-square-trinomial.html .
Guidance
We use the square of a binomial formula to factor perfect square trinomials. A perfect square trinomial has the form
a2 + 2ab + b2 or a2 − 2ab + b2 .
In these special kinds of trinomials, the first and last terms are perfect squares and the middle term is twice the
product of the square roots of the first and last terms. In a case like this, the polynomial factors into perfect squares:
a2 + 2ab + b2 = (a + b)2
a2 − 2ab + b2 = (a − b)2
Once again, the key is figuring out what the a and b terms are.
Example A
Factor the following perfect square trinomials:
a) x2 + 8x + 16
b) x2 − 4x + 4
c) x2 + 14x + 49
Solution
1
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a) The first step is to recognize that this expression is a perfect square trinomial.
First, we can see that the first term and the last term are perfect squares. We can rewrite x2 + 8x + 16 as x2 + 8x + 42 .
Next, we check that the middle term is twice the product of the square roots of the first and the last terms. This is
true also since we can rewrite x2 + 8x + 16 as x2 + 2 · 4 · x + 42 .
This means we can factor x2 + 8x + 16 as (x + 4)2 . We can check to see if this is correct by multiplying (x + 4)2 =
(x + 4)(x + 4) :
x+4
x+4
4x + 16
x2 + 4x
x2 + 8x + 16
The answer checks out.
Note: We could factor this trinomial without recognizing it as a perfect square. We know that a trinomial factors as
a product of two binomials:
(x )(x )
We need to find two numbers that multiply to 16 and add to 8. We can write 16 as the following products:
16 = 1 · 16
and
1 + 16 = 17
16 = 2 · 8
and
2 + 8 = 10
16 = 4 · 4
and
4+4 = 8
T hese are the correct numbers
So we can factor x2 + 8x + 16 as (x + 4)(x + 4), which is the same as (x + 4)2 .
Once again, you can factor perfect square trinomials the normal way, but recognizing them as perfect squares gives
you a useful shortcut.
b) Rewrite x2 + 4x + 4 as x2 + 2 · (−2) · x + (−2)2 .
We notice that this is a perfect square trinomial, so we can factor it as (x − 2)2 .
c) Rewrite x2 + 14x + 49 as x2 + 2 · 7 · x + 72 .
We notice that this is a perfect square trinomial, so we can factor it as (x + 7)2 .
Example B
Factor the following perfect square trinomials:
a) 4x2 + 20x + 25
b) 9x2 − 24x + 16
c) x2 + 2xy + y2
Solution
2
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Chapter 1. Factorization using Perfect Square Trinomials
a) Rewrite 4x2 + 20x + 25 as (2x)2 + 2 · 5 · (2x) + 52 .
We notice that this is a perfect square trinomial and we can factor it as (2x + 5)2 .
b) Rewrite 9x2 − 24x + 16 as (3x)2 + 2 · (−4) · (3x) + (−4)2 .
We notice that this is a perfect square trinomial and we can factor it as (3x − 4)2 .
We can check to see if this is correct by multiplying (3x − 4)2 = (3x − 4)(3x − 4):
3x − 4
3x − 4
− 12x + 16
2
9x − 12x
9x2 − 24x + 16
The answer checks out.
c) x2 + 2xy + y2
We notice that this is a perfect square trinomial and we can factor it as (x + y)2 .
Solve Quadratic Polynomial Equations by Factoring
With the methods we’ve learned in the last two sections, we can factor many kinds of quadratic polynomials. This
is very helpful when we want to solve them. Remember the process we learned earlier:
1.
2.
3.
4.
5.
If necessary, rewrite the equation in standard form so that the right-hand side equals zero.
Factor the polynomial completely.
Use the zero-product rule to set each factor equal to zero.
Solve each equation from step 3.
Check your answers by substituting your solutions into the original equation
We can use this process to solve quadratic polynomials using the factoring methods we just learned.
Example C
Solve the following polynomial equations.
a) x2 + 12x + 36 = 0
b) x2 − 24x = −144
Solution
a) Rewrite:
The equation is already in the correct form.
Factor:
Rewrite x2 + 12x + 36 = 0 as x2 + 2(6x) + 62 = 0. We notice that this is a perfect square trinomial and we can factor
it as (x + 6)2 .
Set the factor equal to zero:
x+6 = 0
3
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Solve:
x = −6
Check: Substitute each solution back into the original equation.
(−6)2 + 12(−6) + 36 =
Substitute in -6.
36 + −72 + 36 =
Simplify.
72 + −72 = 0
Checks out.
b) Rewrite: x2 − 24x = −144 is rewritten as x2 − 24x + 144 = 0
Factor:
x2 − 24x + 144 = x2 + 2(−12)x + (−12)2 = (x − 12)2
Set the factor equal to zero:
x − 12 = 0
Solve:
x = 12
Check: Substitute the solution back into the original equation.
(12)2 − 24(12) + 144 =
Substitute in 12.
144 − 288 + 144 =
Simplify.
288 − 288 = 0
Checks out.
Watch this video for help with the Examples above.
MEDIA
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CK-12 Foundation: Factoring Perfect Square Trinomials
Vocabulary
• A perfect square trinomial has the form
a2 + 2ab + b2 = (a + b)2
4
or
a2 − 2ab + b2 = (a − b)2 .
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Chapter 1. Factorization using Perfect Square Trinomials
Guided Practice
Solve the following polynomial equations:
a) x2 + x + 0.25 = 0
b) x2 − 81 = 0
Solution
a) x2 + x + 0.25 = 0
Rewrite: The equation is in the correct form already.
Factor: Rewrite x2 + x + 0.25 = 0 as x2 + 2 · (0.5)x + (0.5)2 .
We recognize this as a perfect square. This factors as (x + 0.5)2 = 0 or (x + 0.5)(x + 0.5) = 0
Set the factor equal to zero:
x + 0.25 = 0
Solve:
x = −0.5
Check: Substitute the solution back into the original equation.
(−0.5)2 + −0.5 + 0.25 =
0.25 + −0.5 + 0.25 =
Substitute in -0.5
Simplify.
0.5 − 0.5 = 0
Checks out.
b) x2 − 81 = 0
Rewrite: this is not necessary since the equation is in the correct form already
Factor: Rewrite x2 − 81 as x2 − 92 .
We recognize this as a difference of squares. This factors as (x − 9)(x + 9) = 0.
Set each factor equal to zero:
x−9 = 0
or
x+9 = 0
Solve:
x=9
or
x = −9
Check: Substitute each solution back into the original equation.
x=9
x = −9
92 − 81 = 81 − 81 = 0
2
(−9) − 81 = 81 − 81 = 0
checks out
checks out
5
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c) x2 + 20x + 100 = 0
Rewrite: this is not necessary since the equation is in the correct form already
Factor: Rewrite x2 + 20x + 100 as x2 + 2 · 10 · x + 102 .
We recognize this as a perfect square. This factors as (x + 10)2 = 0 or (x + 10)(x + 10) = 0
Set each factor equal to zero:
x + 10 = 0
x + 10 = 0
or
Solve:
x = −10
or
x = −10
This is a double root.
Check: Substitute each solution back into the original equation.
x = 10
(−10)2 + 20(−10) + 100 = 100 − 200 + 100 = 0
Explore More
Factor the following perfect square trinomials.
1.
2.
3.
4.
5.
6.
7.
8.
x2 + 8x + 16
x2 − 18x + 81
−x2 + 24x − 144
x2 + 14x + 49
4x2 − 4x + 1
25x2 + 60x + 36
4x2 − 12xy + 9y2
x4 + 22x2 + 121
Solve the following quadratic equations using factoring.
9.
10.
11.
12.
13.
14.
15.
16.
6
x2 − 11x + 30 = 0
x2 + 4x = 21
x2 + 49 = 14x
x2 − 64 = 0
x2 − 24x + 144 = 0
4x2 − 25 = 0
x2 + 26x = −169
−x2 − 16x − 60 = 0
checks out