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## Алгебра 1

### Course: Алгебра 1>Unit 13

Lesson 7: Factoring quadratics with perfect squares

# Квадраттарды жіктеу: Идеал квадраттар

Learn how to factor quadratics that have the "perfect square" form. For example, write x²+6x+9 as (x+3)².
Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication.
In this article, we'll learn how to factor perfect square trinomials using special patterns. This reverses the process of squaring a binomial, so you'll want to understand that completely before proceeding.

## Intro: Factoring perfect square trinomials

To expand any binomial, we can apply one of the following patterns.
• $\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}$
• $\left(a-b{\right)}^{2}={a}^{2}-2ab+{b}^{2}$
Note that in the patterns, $a$ and $b$ can be any algebraic expression. For example, suppose we want to expand $\left(x+5{\right)}^{2}$. In this case, $a=x$ and $b=5$, and so we get:
$\begin{array}{rl}\left(x+5{\right)}^{2}& ={x}^{2}+2\left(x\right)\left(5\right)+\left(5{\right)}^{2}\\ \\ & ={x}^{2}+10x+25\end{array}$
You can check this pattern by using multiplication to expand $\left(x+5{\right)}^{2}$.
The reverse of this expansion process is a form of factoring. If we rewrite the equations in the reverse order, we will have patterns for factoring polynomials of the form ${a}^{2}±2ab+{b}^{2}$.
We can apply the first pattern to factor ${x}^{2}+10x+25$. Here we have $a=x$ and $b=5$.
$\begin{array}{rl}{x}^{2}+10x+25& ={x}^{2}+2\left(x\right)\left(5\right)+\left(5{\right)}^{2}\\ \\ & =\left(x+5{\right)}^{2}\end{array}$
Expressions of this form are called perfect square trinomials. The name reflects the fact that this type of three termed polynomial can be expressed as a perfect square!
Let's take a look at a few examples in which we factor perfect square trinomials using this pattern.

## Example 1: Factoring ${x}^{2}+8x+16$‍

Notice that both the first and last terms are perfect squares: ${x}^{2}=\left(x{\right)}^{2}$ and $16=\left(4{\right)}^{2}$. Additionally, notice that the middle term is two times the product of the numbers that are squared: $2\left(x\right)\left(4\right)=8x$.
This tells us that the polynomial is a perfect square trinomial, and so we can use the following factoring pattern.
In our case, $a=x$ and $b=4$. We can factor our polynomial as follows:
$\begin{array}{rl}{x}^{2}+8x+16& =\left(x{\right)}^{2}+2\left(x\right)\left(4\right)+\left(4{\right)}^{2}\\ \\ & =\left(x+4{\right)}^{2}\end{array}$
We can check our work by expanding $\left(x+4{\right)}^{2}$:
$\begin{array}{rl}\left(x+4{\right)}^{2}& =\left(x{\right)}^{2}+2\left(x\right)\left(4\right)+\left(4{\right)}^{2}\\ \\ & ={x}^{2}+8x+16\end{array}$

### Тақырып бойынша біліміңді тексер

1) Factor ${x}^{2}+6x+9$.
Дұрыс жауапты таңдаңыз:

2) Factor ${x}^{2}-6x+9$.
Дұрыс жауапты таңдаңыз:

3) Factor ${x}^{2}+14x+49$.

## Example 2: Factoring $4{x}^{2}+12x+9$‍

It is not necessary for the leading coefficient of a perfect square trinomial to be $1$.
For example, in $4{x}^{2}+12x+9$, notice that both the first and last terms are perfect squares: $4{x}^{2}=\left(2x{\right)}^{2}$ and $9=\left(3{\right)}^{2}$. Additionally, notice that the middle term is two times the product of the numbers that are squared: $2\left(2x\right)\left(3\right)=12x$.
Because it satisfies the above conditions, $4{x}^{2}+12x+9$ is also a perfect square trinomial. We can again apply the following factoring pattern.
In this case, $a=2x$ and $b=3$. The polynomial factors as follows:
$\begin{array}{rl}4{x}^{2}+12x+9& =\left(2x{\right)}^{2}+2\left(2x\right)\left(3\right)+\left(3{\right)}^{2}\\ \\ & =\left(2x+3{\right)}^{2}\end{array}$
We can check our work by expanding $\left(2x+3{\right)}^{2}$.

4) Factor $9{x}^{2}+30x+25$.
5) Factor $4{x}^{2}-20x+25$.
6*) Factor ${x}^{4}+2{x}^{2}+1$.
7*) Factor $9{x}^{2}+24xy+16{y}^{2}$.