# How do you factor 7x^2+5x-2?

Apr 24, 2018

$\left(7 x - 2\right) \left(x + 1\right)$

#### Explanation:

$7 {x}^{2}$ only has $7 x \mathmr{and} x$ as factors
-2 is $2 \times 1$ with one of them negative.

Apr 24, 2018

$\left(x + 1\right) \left(7 x - 2\right)$

#### Explanation:

Multiply the coefficient by the constant:

$7 \times - 2 = - 14$

Replace it as the new constant:

$\to 7 {x}^{2} + 5 x - 14$

Find the numbers:

$7 - 2 = 5$

$7 \times - 2 = - 14$

So these work.

Substitute these in:

$7 {x}^{2} + 7 x - 2 x - 2$ $\textcolor{b l u e}{\Leftarrow \text{Notice we replaced the 14 with 2}}$

Factorise the first two expressions:

$7 {x}^{2} + 7 x \to 7 x \left(x + 1\right)$

Factorise the next two expressions:

$- 2 x - 2$

$- 2 \left(x + 1\right)$ $\textcolor{b l u e}{\Leftarrow \text{Notice the brackets are the same}}$

Therefore we get:

$\textcolor{red}{7 x} \textcolor{g r e e n}{\left(x + 1\right)} \textcolor{red}{- 2} \textcolor{g r e e n}{\left(x + 1\right)}$

One bracket is $\left(x + 1\right)$ and the other bracket is the remaining terms:

$\to \left(7 x - 2\right)$

We can always check by expanding:

$\left(7 x - 2\right) \left(x + 1\right) = 7 {x}^{2} + 7 x - 2 x - 2$

$\to 7 {x}^{2} + 5 x - 2$

Which is the original, so we are correct.