Бұл хабарлама біздің веб-сайтқа сыртқы ресурстарды жүктеу кезінде қиындықтар туындағанын білдіреді.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Негізгі бет

### Course: Алгебра 1>Unit 13

Lesson 6: Factoring quadratics with difference of squares

# Квадраттық теңдеулерді жіктеу: Квадраттардың айырымы

Learn how to factor quadratics that have the "difference of squares" form. For example, write x²-16 as (x+4)(x-4).
Көпмүшелерді көбейткіштерге жіктеу үшін оларды екі немесе одан көп көпмүшелердің көбейтіндісі түріне келтіру керек. Бұл көпмүшелерді көбейтуге кері амал болып табылады.
In this article, we'll learn how to use the difference of squares pattern to factor certain polynomials. If you don't know the difference of squares pattern, please check out our video before proceeding.

## Intro: Difference of squares pattern

Every polynomial that is a difference of squares can be factored by applying the following formula:
${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$
Note that $a$ and $b$ in the pattern can be any algebraic expression. For example, for $a=x$ and $b=2$, we get the following:
$\begin{array}{r}{x}^{2}-{2}^{2}=\left(x+2\right)\left(x-2\right)\end{array}$
The polynomial ${x}^{2}-4$ is now expressed in factored form, $\left(x+2\right)\left(x-2\right)$. We can expand the right-hand side of this equation to justify the factorization:
$\begin{array}{rl}\left(x+2\right)\left(x-2\right)& =x\left(x-2\right)+2\left(x-2\right)\\ \\ & ={x}^{2}-2x+2x-4\\ \\ & ={x}^{2}-4\end{array}$
Now that we understand the pattern, let's use it to factor a few more polynomials.

## Example 1: Factoring ${x}^{2}-16$‍

Both ${x}^{2}$ and $16$ are perfect squares, since ${x}^{2}=\left(x{\right)}^{2}$ and $16=\left(4{\right)}^{2}$. In other words:
${x}^{2}-16=\left(x{\right)}^{2}-\left(4{\right)}^{2}$
Since the two squares are being subtracted, we can see that this polynomial represents a difference of squares. We can use the difference of squares pattern to factor this expression:
${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$
In our case, $a=x$ and $b=4$. Therefore, our polynomial factors as follows:
$\left(x{\right)}^{2}-\left(4{\right)}^{2}=\left(x+4\right)\left(x-4\right)$
We can check our work by ensuring the product of these two factors is ${x}^{2}-16$.

### Тақырып бойынша біліміңді тексер

1) Factor ${x}^{2}-25$.
Дұрыс жауапты таңдаңыз:

2) Factor ${x}^{2}-100$.

### Ойлануға арналған сұрақ

3) Can we use the difference of squares pattern to factor ${x}^{2}+25$?
Дұрыс жауапты таңдаңыз:

## Example 2: Factoring $4{x}^{2}-9$‍

The leading coefficient does not have to equal to $1$ in order to use the difference of squares pattern. In fact, the difference of squares pattern can be used here!
This is because $4{x}^{2}$ and $9$ are perfect squares, since $4{x}^{2}=\left(2x{\right)}^{2}$ and $9=\left(3{\right)}^{2}$. We can use this information to factor the polynomial using the difference of squares pattern:
$\begin{array}{rl}4{x}^{2}-9& =\left(2x{\right)}^{2}-\left(3{\right)}^{2}\\ \\ & =\left(2x+3\right)\left(2x-3\right)\end{array}$
A quick multiplication check verifies our answer.

### Тақырып бойынша біліміңді тексер

4) Factor $25{x}^{2}-4$.
Дұрыс жауапты таңдаңыз:

5) Factor $64{x}^{2}-81$.

6) Factor $36{x}^{2}-1$.

## Күрделі есептер

7*) Factor ${x}^{4}-9$.

8*) Factor $4{x}^{2}-49{y}^{2}$.