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# Factor Polynomials Using Special Products

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Practice Factor Polynomials Using Special Products
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# Special Cases of Quadratic Factorization

A box is to be designed for packaging with a side length represented by the quadratic $9b^2 - 64$ . If this is the most economical box, what are the dimensions?

### Guidance

When factoring quadratics, there are special cases that can be factored more quickly. There are two special quadratics that you should learn to recognize:

Special Case 1 (Perfect Square Trinomial): $x^2 \pm 2xy + y^2=(x\pm y)^2$

• Example: $x^2 + 10x +25=(x+5)^2$
• Example: $4x^2 -32x + 64=(2x-8)^2$

Special Case 2 (Difference of Perfect Squares): $x^2 - y^2=(x+y)(x-y)$

• Example: $25x^2 - 100=(5x+10)(5x-10)$
• Example: $4x^2-25=(2x-5)(2x+5)$

Keep in mind that you can always use the box method to do the factoring if you don't notice the problem as a special case.

#### Example A

Factor $2x^2+28x+98$ .

Solution: First, notice that there is a common factor of 2. Factor out the common factor:

$2x^2+28x+98=2(x^2+14x+49)$

Next, notice that the first and last terms are both perfect squares ( $x^2=x\cdot x$ and $49=7\cdot 7$ , and the middle term is 2 times the product of the roots of the other terms ( $14x=2\cdot x\cdot 7$ ). This means $x^2+14x+49$ is a perfect square trinomial (Special Case 1). Using the pattern:

$x^2+14x+49=(x+7)^2$

Therefore, the complete factorization is $2x^2+28x+98=2(x+7)^2$ .

#### Example B

Factor $8a^2-24a+18$ .

Solution: First, notice that there is a common factor of 2. Factor out the common factor:

$8a^2-24a+18=2(4a^2-12a+9)$

Next, notice that the first and last terms are both perfect squares and the middle term is 2 times the product of the roots of the other terms ( $12a=2\cdot 2a\cdot 3$ ). This means $4a^2-12a+9$ is a perfect square trinomial (Special Case 1). Because the middle term is negative, there will be a negative in the binomial. Using the pattern:

$4a^2-12a+9=(2a-3)^2$

Therefore, the complete factorization is $8a^2-24a+18=2(2a-3)^2$ .

#### Example C

Factor $x^2-16$ .

Solution: Notice that there are no common factors. The typical middle term of the quadratic is missing and each of the terms present are perfect squares and being subtracted. This means $x^2-16$ is a difference of perfect squares (Special Case 2). Using the pattern:

$x^2-16=(x-4)(x+4)$

Note that it would also be correct to say $x^2-16=(x+4)(x-4)$ . It does not matter whether you put the + version of the binomial first or the – version of the binomial first.

#### Concept Problem Revisited

A box is to be designed for packaging with a side length represented by the quadratic $9b^2 - 64$ . If this is the most economical box, what are the dimensions?

First: factor the quadratic to find the value for $b$ .

$9b^2-64$

This is a difference of perfect squares (Special Case 2). Use that pattern:

$9b^2-64=(3b-8)(3b+8)$

To finish this problem we need to solve a quadratic equation. This idea is explored in further detail in another concept.

$& 9b^2-64=(3b+8)(3b-8)\\& \qquad \qquad \quad \swarrow \qquad \qquad \searrow\\& \ \quad 3b+8=0 \qquad \qquad 3b-8=0\\& \ \qquad \ \ 3b=-8 \qquad \qquad \quad 3b=8\\& \ \qquad \quad b=\frac{-8}{3} \qquad \qquad \quad \ b=\frac{8}{3}$

The most economical box is a cube. Therefore the dimensions are $\frac{8}{3} \times \frac{8}{3} \times \frac{8}{3}$

### Vocabulary

Difference of Perfect Squares
The difference of perfect squares is a special case of a quadratic expression where there is no middle term and the two terms present are both perfect squares. The general equation for the difference of two squares is:

$x^2-y^2=(x+y)(x-y)$

Perfect Square Trinomial
The perfect square trinomials are the result of a binomial being multiplied by itself. The two variations of the perfect square trinomial are:
1. $(x+y)^2=x^2+2xy+y^2$
2. $(x-y)^2=x^2-2xy+y^2$

### Guided Practice

1. Factor completely $s^2-18s+81$

2. Factor completely $50-98x^2$

3. Factor completely $4x^2+48x+144$

1. This is Special Case 1. $s^2-18s+81=(s-9)^2$

2. First factor out the common factor of 2. Then, it is Special Case 2. $50-98x^2=2(5-7x)(5+7x)$

3. First factor out the common factor of 4. Then, it is Special Case 1. $4x^2+48x+144=4(x+6)^2$

### Practice

Factor each of the following:

1. $s^2+18s+81$
2. $x^2+12x+36$
3. $y^2-14y+49$
4. $4a^2+20a+25$
5. $9s^2-48s+64$
6. $s^2-81$
7. $x^2-49$
8. $4t^2-25$
9. $25w^2-36$
10. $64-81a^2$
11. $y^2-22y+121$
12. $16t^2-49$
13. $9a^2+30a+25$
14. $100-25b^2$
15. $4s^2-28s+49$