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### Course: Алгебра 1>Unit 13

Lesson 4: Квадраттық теңдеулерді көбейткішке жіктеуге кіріспе

Learn how to factor quadratic expressions as the product of two linear binomials. For example, x²+5x+6=(x+2)(x+3).

#### What you need to know for this lesson

Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication. For more on this, check out our previous article on taking common factors.

#### Бұл сабақта сен нені үйренесің

Бұл сабақта ${x}^{2}+bx+c$ түріндегі көпмүшені екі көпмүшенің көбейтіндісіне түрінде жіктеуді үйренесің.

## Қайталау: көпмүшелерді көбейту

$\left(x+2\right)\left(x+4\right)$ өрнегін қарастырайық.
We can find the product by applying the distributive property multiple times.
Сонымен $\left(x+2\right)\left(x+4\right)={x}^{2}+6x+8$.
From this, we see that $x+2$ and $x+4$ are factors of ${x}^{2}+6x+8$, but how would we find these factors if we didn't start with them?

## Квадрат үшмүшені көбейткіштерге жіктеу

We can reverse the process of binomial multiplication shown above in order to factor a trinomial (which is a polynomial with $3$ terms).
In other words, if we start with the polynomial ${x}^{2}+6x+8$, we can use factoring to write it as a product of two binomials, $\left(x+2\right)\left(x+4\right)$.
Let's take a look at a few examples to see how this is done.

### Example 1: Factoring ${x}^{2}+5x+6$‍

To factor ${x}^{2}+5x+6$, we first need to find two numbers that multiply to $6$ (the constant number) and add up to $5$ (the $x$-coefficient).
These two numbers are $2$ and $3$ since $2\cdot 3=6$ and $2+3=5$.
We can then add each of these numbers to $x$ to form the two binomial factors: $\left(x+2\right)$ and $\left(x+3\right)$.
In conclusion, we factored the trinomial as follows:
${x}^{2}+5x+6=\left(x+2\right)\left(x+3\right)$
To check the factorization, we can multiply the two binomials:
$\begin{array}{rl}\left(x+2\right)\left(x+3\right)& =\left(x+2\right)\left(x\right)+\left(x+2\right)\left(3\right)\\ \\ & ={x}^{2}+2x+3x+6\\ \\ & ={x}^{2}+5x+6\end{array}$
The product of $x+2$ and $x+3$ is indeed ${x}^{2}+5x+6$. Our factorization is correct!

### Түсінгеніңізді тексеріңіз

1) Factor ${x}^{2}+7x+10$.
Дұрыс жауапты таңдаңыз:

2) Factor ${x}^{2}+9x+20$.

Let's take a look at a few more examples and see what we can learn from them.

### 2-мысал: ${x}^{2}-5x+6$‍  өрнегін көбейткіштерге жіктеу

To factor ${x}^{2}-5x+6$, let's first find two numbers that multiply to $6$ and add up to $-5$.
These two numbers are $-2$ and $-3$ since $\left(-2\right)\cdot \left(-3\right)=6$ and $\left(-2\right)+\left(-3\right)=-5$.
We can then add each of these numbers to $x$ to form the two binomial factors: $\left(x+\left(-2\right)\right)$ and $\left(x+\left(-3\right)\right)$.
The factorization is given below:
$\begin{array}{rl}{x}^{2}-5x+6& =\left(x+\left(-2\right)\right)\left(x+\left(-3\right)\right)\\ \\ & =\left(x-2\right)\left(x-3\right)\end{array}$
Factoring pattern: Notice that the numbers needed to factor ${x}^{2}-5x+6$ are both negative $\left(-2$ and $-3\right)$. This is because their product needs to be positive $\left(6\right)$ and their sum negative $\left(-5\right)$.
In general, when factoring ${x}^{2}+bx+c$, if $c$ is positive and $b$ is negative, then both factors will be negative!

### 3-мысал: ${x}^{2}-x-6$‍  өрнегін көбейткіштерге жіктеу

We can write ${x}^{2}-x-6$ as ${x}^{2}-1x-6$.
To factor ${x}^{2}-1x-6$, let's first find two numbers that multiply to $-6$ and add up to $-1$.
These two numbers are $2$ and $-3$ since $\left(2\right)\cdot \left(-3\right)=-6$ and $2+\left(-3\right)=-1$.
We can then add each of these numbers to $x$ to form the two binomial factors: $\left(x+2\right)$ and $\left(x+\left(-3\right)\right)$.
The factorization is given below:
$\begin{array}{rl}{x}^{2}-x-6& =\left(x+2\right)\left(x+\left(-3\right)\right)\\ \\ & =\left(x+2\right)\left(x-3\right)\end{array}$
Factoring patterns: Notice that to factor ${x}^{2}-x-6$, we need one positive number $\left(2\right)$ and one negative number $\left(-3\right)$. This is because their product needs to be negative $\left(-6\right)$.
In general, when factoring ${x}^{2}+bx+c$, if $c$ is negative, then one factor will be positive and one factor will be negative.

## Қорытынды

In general, to factor a trinomial of the form ${x}^{2}+bx+c$, we need to find factors of $c$ that add up to $b$.
Suppose these two numbers are $m$ and $n$ so that $c=mn$ and $b=m+n$, then ${x}^{2}+bx+c=\left(x+m\right)\left(x+n\right)$.

### Түсінгеніңізді тексеріңіз

3) Көбейткіштерге жіктеңіз: ${x}^{2}-8x-9$.

4) ${x}^{2}-10x+24$ өрнегін көбейткіштерге жіктеңіз.

5) Көбейткіштерге жіктеңіз: ${x}^{2}+7x-30$.

## Why does this work?

To understand why this factorization method works, let's return to the original example in which we factored ${x}^{2}+5x+6$ as $\left(x+2\right)\left(x+3\right)$.
If we go back and multiply the two binomial factors, we can see the effect that the $2$ and the $3$ have on forming the product ${x}^{2}+5x+6$.
$\begin{array}{rl}\left(x+2\right)\left(x+3\right)& =\left(x+2\right)\left(x\right)+\left(x+2\right)\left(3\right)\\ \\ & ={x}^{2}+2x+3x+2\cdot 3\\ \\ & ={x}^{2}+\left(2+3\right)x+2\cdot 3\end{array}$
We see that the coefficient of the $x$-term is the sum of $2$ and $3$, and the constant term is the product of $2$ and $3$.

## The sum-product pattern

$\left(x+m\right)\left(x+n\right)$ үшін $\left(x+2\right)\left(x+3\right)$- мен не жасағанымызды есімізге түсірейік:
$\begin{array}{rl}\left(x+m\right)\left(x+n\right)& =\left(x+m\right)\left(x\right)+\left(x+m\right)\left(n\right)\\ \\ & ={x}^{2}+mx+nx+m\cdot n\\ \\ & ={x}^{2}+\left(m+n\right)x+m\cdot n\end{array}$
To summarize this process, we get the following equation:
$\left(x+m\right)\left(x+n\right)={x}^{2}+\left(m+n\right)x+m\cdot n$
This is called the sum-product pattern.
It shows why, once we express a trinomial ${x}^{2}+bx+c$ as ${x}^{2}+\left(m+n\right)x+m\cdot n$ (by finding two numbers $m$ and $n$ so $b=m+n$ and $c=m\cdot n$), we can factor that trinomial as $\left(x+m\right)\left(x+n\right)$.

### Reflection question

6) Can this factorization method be used to factor $2{x}^{2}+3x+1$?
Дұрыс жауапты таңдаңыз:

## When can we use this method to factor?

In general, the sum-product method is only applicable when we can actually write a trinomial as $\left(x+m\right)\left(x+n\right)$ for some integers $m$ and $n$.
This means that the leading term of the trinomial must be ${x}^{2}$ (and not, for instance, $2{x}^{2}$) in order to even consider this method. This is because the product of $\left(x+m\right)$ and $\left(x+n\right)$ will always be a polynomial with a leading term of ${x}^{2}$.
However, not all trinomials with ${x}^{2}$ as a leading term can be factored. For example, ${x}^{2}+2x+2$ cannot be factored because there are no two integers whose sum is $2$ and whose product is $2$.
In future lessons we will learn more ways of factoring more types of polynomials.

## Күрделі есептер

7*) Көбейткіштерге жіктеңіз: ${x}^{2}+5xy+6{y}^{2}$.

8*) Көбейткіштерге жіктеңіз: ${x}^{4}-5{x}^{2}+6$