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# Special Cases - Squares

### Learning Outcomes

• Factor special products
Some people find it helpful to know when they can take a shortcut to avoid doing extra work. There are some polynomials that will always factor a certain way, and for those, we offer a shortcut. Most people find it helpful to memorize the factored form of a perfect square trinomial or a difference of squares. The most important skill you will use in this section will be recognizing when you can use the shortcuts.

## Factoring a Perfect Square Trinomial

A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.
$\begin{array}{ccc}\hfill {a}^{2}+2ab+{b}^{2}& =& {\left(a+b\right)}^{2}\hfill \\ & \text{and}& \\ \hfill {a}^{2}-2ab+{b}^{2}& =& {\left(a-b\right)}^{2}\hfill \end{array}$
We can use these equations to factor any perfect square trinomial.

### A General Note: Perfect Square Trinomials

A perfect square trinomial can be written as the square of a binomial:
${a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}$
${a}^{2}-2ab+{b}^{2}={\left(a-b\right)}^{2}$
In the following example, we will show you how to define a and b so you can use the shortcut.

### Example

Factor $25{x}^{2}+20x+4$.

Answer: First, notice that $25{x}^{2}$ and $4$ are perfect squares because $25{x}^{2}={\left(5x\right)}^{2}$ and $4={2}^{2}$. This means that $a=5x\text{ and }b=2$ Next, check to see if the middle term is equal to $2ab$, which it is:

$2ab = 2\left(5x\right)\left(2\right)=20x$

Therefore, the trinomial is a perfect square trinomial and can be written as ${\left(a+b\right)}^{2}={\left(5x+2\right)}^{2}$.

In the next example, we will show that we can use $1 = 1^2$ to factor a polynomial with a term equal to $1$.

### Example

Factor $49{x}^{2}-14x+1$.

Answer: First, notice that $49{x}^{2}$ and $1$ are perfect squares because $49{x}^{2}={\left(7x\right)}^{2}$ and $1={1}^{2}$. This means that $a=7x$ and $b=1$. Next, check to see if the middle term is equal to $2ab$, which it is:

$2ab = 2\left(7x\right)\left(1\right)=14x$

Therefore, the trinomial is a perfect square trinomial and can be written as ${\left(a-b\right)}^{2}={\left(7x-1\right)}^{2}$.

In the following video, we provide another short description of what a perfect square trinomial is and show how to factor them using a formula. https://youtu.be/UMCVGDTxxTI

### Try It

[ohm_question]91970[/ohm_question]
We can summarize our process in the following way:

### How To: Given a perfect square trinomial, factor it into the square of a binomial

1. Confirm that the first and last term are perfect squares.
2. Confirm that the middle term is twice the product of $ab$.
3. Write the factored form as ${\left(a+b\right)}^{2}$ or ${\left(a-b\right)}^{2}$.

## Factoring a Difference of Squares

A difference of squares is a perfect square subtracted from a perfect square. This type of polynomial is unique because it can be factored into two binomials but has only two terms.

### Factor a Difference of Squares

Given $a^2-b^2$, its factored form will be $\left(a+b\right)\left(a-b\right)$.
You will want to become familiar with the special relationship between a difference of squares and its factorization as we can use this equation to factor any differences of squares. A difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.  Let's look at an example of difference of squares to help us understand how this works.  We will start from the product of two binomials to see the pattern. Given the product of two binomials: $\left(x-2\right)\left(x+2\right)$, if we multiply them together, we lose the middle term that we are used to seeing as a result.

Multiply:

$\begin{array}{l}\left(x-2\right)\left(x+2\right)\\\text{}\\=x^2-2x+2x-2^2\\\text{}\\=x^2-2^2\\\text{}\\=x^2-4\end{array}$

The polynomial $x^2-4$ is called a difference of squares because each term can be written as something squared.  A difference of squares will always factor in the following way: Let’s factor $x^{2}–4$ by writing it as a trinomial, $x^{2}+0x–4$.  This is similar in format to the trinomials we have been factoring so far, so let’s use the same method.

Find the factors of $a\cdot{c}$ whose sum is b, in this case, 0:

Factors of $−4$ Sum of the factors
$1\cdot-4=−4$ $1-4=−3$
$2\cdot−2=−4$ $2-2=0$
$-1\cdot4=−4$ $-1+4=3$
$2$, and $-2$ have a sum of $0$. You can use these to factor $x^{2}–4$.

### Example

Factor $x^{2}–4$.

Answer: Rewrite $0x$ as $−2x+2x$.

$\begin{array}{l}x^{2}+0x-4\\x^{2}-2x+2x-4\end{array}$

Group pairs.

$\left(x^{2}–2x\right)+\left(2x–4\right)$

Factor x out of the first group. Factor $2$ out of the second group.

$x\left(x–2\right)+2\left(x–2\right)$

Factor out $\left(x–2\right)$.

$\left(x–2\right)\left(x+2\right)$

[latex-display]\left(x–2\right)\left(x+2\right)[/latex-display]

Since order doesn't matter with multiplication, the answer can also be written as $\left(x+2\right)\left(x–2\right)$. You can check the answer by multiplying $\left(x–2\right)\left(x+2\right)=x^{2}+2x–2x–4=x^{2}–4$.

### A General Note: Differences of Squares

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.
${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$
Now that we have seen how to factor a difference of squares with regrouping, let's try some examples using the short-cut.

### Example

Factor $9{x}^{2}-25$.

Answer: Notice that $9{x}^{2}$ and $25$ are perfect squares because $9{x}^{2}={\left(3x\right)}^{2}$ and $25={5}^{2}$.This means that $a=3x,\text{ and }b=5$ The polynomial represents a difference of squares and can be rewritten as $\left(3x+5\right)\left(3x - 5\right)$. Check that you are correct by multiplying. [latex-display]\left(3x+5\right)\left(3x - 5\right)=9x^2-15x+15x-25=9x^2-25[/latex-display]

### Try It

[ohm_question]161674[/ohm_question]
The most helpful thing for recognizing a difference of squares that can be factored with the shortcut is knowing which numbers are perfect squares, as you will see in the next example.

### Example

Factor $81{y}^{2}-144$.

Answer: Notice that $81{y}^{2}$ and $144$ are perfect squares because $81{y}^{2}={\left(9x\right)}^{2}$ and $144={12}^{2}$. This means that $a=9x,\text{ and }b=12$ The polynomial represents a difference of squares and can be rewritten as $\left(9x+12\right)\left(9x - 12\right)$. Check that you are correct by multiplying. [latex-display]\left(9x+12\right)\left(9x - 12\right)=81x^2-108x+108x-144=81x^2-144[/latex-display]

TIP To help identify difference of squares factoring problems, make a list of perfect squares and become familiar with these values.  You will frequently see them in these types of factoring problems.  In addition, these problems must have a minus sign. For example, x^2 - 9 = (x+3)(x-3)
 1^2 1 2^2 4 3^2 9 4^2 16 5^2 25

### Try It

[ohm_question]7930[/ohm_question]
In the following video, we show another example of how to use the formula for factoring a difference of squares. https://youtu.be/Li9IBp5HrFA We can summarize the process for factoring a difference of squares with the shortcut this way:

### How To: Given a difference of squares, factor it into binomials

1. Confirm that the first and last term are perfect squares.
2. Write the factored form as $\left(a+b\right)\left(a-b\right)$.

Is there a formula to factor the sum of squares, $a^2+b^2$, into a product of two binomials? Write down some ideas for how you would answer this in the box below before you look at the answer. [practice-area rows="1"][/practice-area]

Answer: There is no way to factor a sum of squares into a product of two binomials. This is because of addition - the middle term needs to "disappear" and the only way to do that is with opposite signs. To get a positive result, you must multiply two numbers with the same signs. The only time a sum of squares can be factored is if they share any common factors, as in the following case: [latex-display]9x^2+36[/latex-display] The only way to factor this expression is by pulling out the GCF which is 9. [latex-display]9x^2+36=9(x^2+4)[/latex-display]

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