<meta http-equiv="refresh" content="1; url=/nojavascript/"> Factoring Special Quadratics | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra II with Trigonometry Concepts Go to the latest version.

# 5.3: Factoring Special Quadratics

Difficulty Level: At Grade / Advanced / Basic Created by: CK-12

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.

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

### Guidance

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.

$4(9x^2+30x+25)$

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. $81x^2+144+64$

#### Answers

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 $b = 5$.

$2(x^2-10x+25)=2(x-5)^2.$

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 $(9x + 8)^2$.

### Vocabulary

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.

### Practice

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:

$(2x-5)^2=(2x)^2-5^2=4x^2-25$

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

At Grade

Mar 12, 2013

## Last Modified:

Dec 13, 2013
Files can only be attached to the latest version of Modality

# Reviews

Please wait...
You need to be signed in to perform this action. Please sign-in and try again.
Please wait...
Image Detail
Sizes: Medium | Original

MAT.ALG.624.1.L.2