# How do you factor: y= x^2 + 7x + 12 ?

Apr 12, 2018

$y = \left(x + 4\right) \left(x + 3\right)$

#### Explanation:

y=x^2+7x+12)

$y = \left(x + 4\right) \left(x + 3\right)$

Apr 12, 2018

$\left(x + 3\right) \left(x + 4\right)$

#### Explanation:

$\text{the factors of + 12 which sum to + 7 are + 3 and + 4}$

$\Rightarrow {x}^{2} + 7 x + 12 = \left(x + 3\right) \left(x + 4\right)$

Apr 12, 2018

$\left(x + 4\right) \left(x + 3\right)$

#### Explanation:

If a quadratic with $3$ terms; one with a coefficient of $x$, for example, ${x}^{2} , 2 {x}^{2} , 3 {x}^{2}$ etc, a value of $x$ and a constant usually the factorised form includes $2$ brackets.

When factorising that has $2$ brackets we need $2$ numbers that add up to make the second term and the same $2$ numbers to multiply to get the second term.

I usually start by listing the factors of the third term which is $12$:

$12$ and $1$

As $12$ and $1$ cannot add or subtract to make $7$ this pair does not work.

$6$ and $2$

As $6$ and $2$ cannot add or subtract to make $7$ this pair does not work.

$4$ and $3$

$4$ and $3$ add to make $7$ so we can use this.

With all of the signs being positive within ${x}^{2} + 7 x + 12$, both $4$ and $3$ have to be both positive.

$\to \left(x + 3\right) \left(x + 4\right)$

Remember you can always expand each term to check:

$\left(\textcolor{red}{x} \textcolor{b l u e}{+ 3}\right) \left(\textcolor{red}{x} \textcolor{b l u e}{+ 4}\right)$

$\textcolor{red}{x} \times \textcolor{red}{x} = \textcolor{l i g h t g r e e n}{{x}^{2}}$

$\textcolor{red}{x} \times \textcolor{b l u e}{4} = \textcolor{red}{4 x}$

$\textcolor{b l u e}{3} \times \textcolor{red}{x} = \textcolor{red}{3 x}$

$\textcolor{b l u e}{3} \times \textcolor{b l u e}{4} = \textcolor{b l u e}{12}$

$\to \textcolor{l i g h t g r e e n}{{x}^{2}} \textcolor{red}{+ 4 x} \textcolor{red}{+ 3 x} \textcolor{b l u e}{+ 12}$

$\to \to \textcolor{l i g h t g r e e n}{{x}^{2}} \textcolor{red}{+ 7 x} \textcolor{b l u e}{+ 12}$

This is the same as what we started with, therefore, it is correctly factorised.

$\to \left(x + 4\right) \left(x + 3\right)$