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5.3: Factoring Special Quadratics

Difficulty Level: At Grade Created by: CK-12
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The total time, in hours, it takes a rower to paddle upstream, turn around and come back to her starting point is 18x^2 = 32 . How long does it take her to make the round trip?

Watch This

First, watch this video.

Khan Academy: U09_L2_TI_we1 Factoring Special Products 1

Then, watch the first part of this video, until about 3:10

James Sousa: Factoring a Difference of Squares


There are a couple of special quadratics that, when factored, have a pattern.

Investigation: Multiplying the Square of a Binomal

1. Rewrite (a+b)^2 as the product of two factors. Expand (a+b)^2 . (a+b)^2=(a+b)(a+b)

2. FOIL your answer from Step 1. This is a perfect square trinomial. a^2+2ab+b^2

3. (a-b)^2 also produces a perfect square trinomial. (a-b)^2 = a^2-2ab+b^2

4. Apply the formula above to factoring 9x^2-12x+4 . First, find a and b .

a^2 &=9x^2, \ b^2=4\\a &=3x, \quad b=2

5. Now, plug a and b into the appropriate formula.

(3x-2)^2 & = (3x)^2 - 2(3x)(2)+2^2\\& = 9x^2-12x+4

Investigation: Multiplying (a + b)(a - b)

1. FOIL (a-b)(a+b) .

(a-b)(a+b) &= a^2+ab-ab-b^2\\&= a^2-b^2

2. This is a difference of squares. The difference of squares will always factor to be (a + b)(a-b) .

3. Apply the formula above to factoring 25x^2-16 . First, find a and b .

a^2 &=25x^2, \ b^2=16\\a &=5x, \quad \ \ b=4

4. Now, plug a and b into the appropriate formula. (5x-4)(5x+4)=(5x)^2-4^2

^{**} It is important to note that if you forget these formulas or do not want to use them, you can still factor all of these quadratics the same way you did in the previous two concepts.

Example A

Factor x^2-81 .

Solution: Using the formula from the investigation above, we need to first find the values of a and b .

& x^2-81 = a^2-b^2\\& a^2=x^2, \ b^2=81\\& \ a=x, \quad \ b=9

Now, plugging x and 9 into the formula, we have x^2-81 = (x-9)(x+9) . To solve for a and b , we found the square root of each number. Recall that the square root is a number that, when multiplied by itself, produces another number. This other number is called a perfect square .

Alternate Method

Rewrite x^2-81 so that the middle term is present. x^2+0x-81

Using the method from the previous two concepts, what are the two factors of -81 that add up to 0? 9 and -9

Therefore, the factors are (x-9)(x + 9) .

Example B

Factor 36x^2+120x+100 .

Solution: First, check for a GCF.


Now, double-check that the quadratic equation above fits into the perfect square trinomial formula.

a^2 &= 9x^2 \qquad \quad \quad b^2 = 25 \\\sqrt{a^2} &= \sqrt{9x^2} \qquad \ \sqrt{b^2} = \sqrt{25} \qquad \quad \quad 2ab=30x\\a &=3x \qquad \qquad \quad b=5 \qquad \quad \ 2(3x)(5)=30x

Using a and b above, the equation factors to be 4(3x + 5)^2 . If you did not factor out the 4 in the beginning, the formula will still work. a would equal 6x and b would equal 10, so the factors would be (6x + 10)^2 . If you expand and find the GCF, you would have (6x+10)^2=(6x+10)(6x+10)=2(3x+5)2(3x+5)=4(3x+5)^2 .

Alternate Method

First, find the GCF. 4(9x^2+30x+25)

Then, find ac and expand b accordingly. 9 \cdot 25 = 225 , the factors of 225 that add up to 30 are 15 and 15.

& 4(9x^2+{\color{blue}30x}+25)\\& 4(9x^2+{\color{blue}15x+15x}+25)\\& 4 \left[(9x^2+15x)+(15x+25) \right]\\& 4 \left[3x(3x+5)+5(3x+5) \right]\\& 4(3x+5)(3x+5) \ or \ 4(3x^2+5)

Again, notice that if you do not use the formula discovered in this concept, you can still factor and get the correct answer.

Example C

Factor 48x^2-147 .

Solution: At first glance, this does not look like a difference of squares. 48 nor 147 are square numbers. But, if we take a 3 out of both, we have 3(16x^2-49) . 16 and 49 are both square numbers, so now we can use the formula.

16x^2 &=a^2 \qquad 49=b^2\\4x &=a \qquad \quad 7=b

The factors are 3(4x-7)(4x+7) .

Intro Problem Revisit 18x^2 = 32 can be rewritten as 18x^2-32 = 0 , so factor 18x^2-32 .

First, we must take greatest common factor of 2 out of both. We then have 2(9x^2-16) . 9 and 16 are both square numbers, so now we can use the formula.

9x^2 &=a^2 \qquad 16=b^2\\3x &=a \qquad \quad 4=b

The factors are 2(3x-4)(3x+4) .

Finally, to find the time, set these factors equal to zero and solve 2(3x-4)(3x+4) = 0 .

Because x represents the time, it must be positive. Only (3x-4) = 0 results in a positive value of x .

x = \frac{4}{3} = 1.3333 Therefore the round trip takes 1.3333 hours.

You will do more problems like this one in the next lesson.

Guided Practice

Factor the following quadratic equations.

1. x^2-4

2. 2x^2-20x+50

3. 81x^2+144+64


1. a = x and b = 2 . Therefore, x^2-4=(x-2)(x+2) .

2. Factor out the GCF, 2. 2(x^2-10x+25) . This is now a perfect square trinomial with a = x and b = 5 .


3. This is a perfect square trinomial and no common factors. Solve for a and b .

81x^2 &=a^2 \qquad 64=b^2\\9x &=a \qquad \quad 8=b

The factors are (9x + 8)^2 .


Perfect Square Trinomial
A quadratic equation in the form a^2+2ab+b^2 or a^2-2ab+b^2 .
Difference of Squares
A quadratic equation in the form a^2-b^2 .
Square Root
A number, that when multiplied by itself produces another number. 3 is the square root of 9.
Perfect Square
A number that has a square root that is an integer. 25 is a perfect square.


  1. List the perfect squares that are less than 200.
  2. Why do you think there is no sum of squares formula?

Factor the following quadratics, if possible.

  1. x^2-1
  2. x^2+4x+4
  3. 16x^2-24x+9
  4. -3x^2+36x-108
  5. 144x^2-49
  6. 196x^2+140x+25
  7. 100x^2+1
  8. 162x^2+72x+8
  9. 225-x^2
  10. 121-132x+36x^2
  11. 5x^2+100x-500
  12. 256x^2-676
  13. Error Analysis Spencer is given the following problem: Multiply (2x-5)^2 . Here is his work:


His teacher tells him the answer is 4x^2-20x+25 . What did Spencer do wrong? Describe his error and correct the problem.

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Difficulty Level:

At Grade


Date Created:

Mar 12, 2013

Last Modified:

May 27, 2014
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