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### Course: Алгебра 1>Unit 13

Lesson 5: Factoring quadratics by grouping

# Factoring by grouping

Learn about a factorization method called "grouping." For example, we can use grouping to write 2x²+8x+3x+12 as (2x+3)(x+4).

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

Көпмүшелерді көбейткіштерге жіктеу үшін оларды екі немесе одан көп көпмүшелердің көбейтіндісі түріне келтіру керек. Бұл көпмүшелерді көбейтуге кері амал болып табылады.
We have seen several examples of factoring already. However, for this article, you should be especially familiar with taking common factors using the distributive property. For example, $6{x}^{2}+4x=2x\left(3x+2\right)$ .

#### Бұл сабақта сен нені үйренесің

In this article, we will learn how to use a factoring method called grouping.

## Example 1: Factoring $2{x}^{2}+8x+3x+12$‍

First, notice that there is no factor common to all terms in $2{x}^{2}+8x+3x+12$. However, if we group the first two terms together and the last two terms together, each group has its own GCF, or greatest common factor:
In particular, there is a GCF of $2x$ in the first grouping and a GCF of $3$ in the second grouping. We can factor these out to obtain the following expression:
$2x\left(x+4\right)+3\left(x+4\right)$
Notice that this reveals yet another common factor between the two terms: $x+4$. We can use the distributive property to factor out this common factor.
Since the polynomial is now expressed as a product of two binomials, it is in factored form. We can check our work by multiplying and comparing it to the original polynomial.

## Example 2: Factoring $3{x}^{2}+6x+4x+8$‍

Let's summarize what was done above by factoring another polynomial.
The factored form is $\left(x+2\right)\left(3x+4\right)$.

### Түсінгеніңізді тексеріңіз

1) Factor $9{x}^{2}+6x+12x+8$.
Дұрыс жауапты таңдаңыз:

2) Factor $5{x}^{2}+10x+2x+4$.

3) Factor $8{x}^{2}+6x+4x+3$.

## Example 3: Factoring $3{x}^{2}-6x-4x+8$‍

Extra care should be taken when using the grouping method to factor a polynomial with negative coefficients.
For example, the steps below can be used to factor $3{x}^{2}-6x-4x+8$.
The factored form of the polynomial is $\left(x-2\right)\left(3x-4\right)$. We can multiply the binomials to check our work.
A few of the steps above may seem different than what you saw in the first example, so you may have a few questions.
Where did the "+" sign between the groupings come from?
In step $\left(1\right)$, a "+" sign was added between the groupings $\left(3{x}^{2}-6x\right)$ and $\left(-4x+8\right)$. This is because the third term $\left(-4x\right)$ is negative, and the sign of the term must be included within the grouping.
Keeping the minus sign outside the second grouping is tricky. For example, a common error is to group $3{x}^{2}-6x-4x+8$ as $\left(3{x}^{2}-6x\right)-\left(4x+8\right)$. This grouping, however, simplifies to $3{x}^{2}-6x-4x-8$, which is not the same as the original expression.
Why factor out $-4$ instead of $4$?
In step $\left(2\right)$, we factored out a $-4$ to reveal a common factor of $\left(x-2\right)$ between the terms. If we instead factored out a positive $4$, we would not obtain that common binomial factor seen above:
$\begin{array}{rl}\left(3{x}^{2}-6x\right)+\left(-4x+8\right)& =3x\left(x-2\right)+4\left(-x+2\right)\end{array}$
When the leading term in a group is negative, we will often need to factor out a negative common factor.

### Түсінгеніңізді тексеріңіз

4) Factor $2{x}^{2}-3x-4x+6$.
Дұрыс жауапты таңдаңыз:

5) Factor $3{x}^{2}+3x-10x-10$.

6) Factor $3{x}^{2}+6x-x-2$.

## Күрделі тапсырма

7*) Factor $2{x}^{3}+10{x}^{2}+3x+15$.

### When can we use the grouping method?

The grouping method can be used to factor polynomials whenever a common factor exists between the groupings.
For example, we can use the grouping method to factor $3{x}^{2}+9x+2x+6$ since it can be written as follows:
$\begin{array}{rl}\left(3{x}^{2}+9x\right)+\left(2x+6\right)& =3x\left(x+3\right)+2\left(x+3\right)\end{array}$
We cannot, however, use the grouping method to factor $2{x}^{2}+3x+4x+12$ because factoring out the GCF from both groupings does not yield a common factor!
$\begin{array}{rl}\left(2{x}^{2}+3x\right)+\left(4x+12\right)& =x\left(2x+3\right)+4\left(x+3\right)\end{array}$

#### Using grouping to factor trinomials

You can also use grouping to factor certain three termed quadratics (i.e. trinomials) like $2{x}^{2}+7x+3$. This is because we can rewrite the expression as follows:
$2{x}^{2}+7x+3=2{x}^{2}+1x+6x+3$
Then we can use grouping to factor $2{x}^{2}+1x+6x+3$ as $\left(x+3\right)\left(2x+1\right)$.
For more on factoring quadratic trinomials like these using the grouping method, check out our next article.