# How do you factor y=2x^2 - 5x – 3 ?

May 7, 2016

$y = \left(2 x + 1\right) \left(x - 3\right)$

#### Explanation:

To find the factors, put $y = 0$

$\implies 2 {x}^{2} - 5 x - 3 = 0$

This is a quadratic equation and will have 2 factors or 2 roots.

We first find two numbers that:

• multiply to $\textcolor{red}{- 6}$ (because $2 \times - 3 = - 6$) and
• add up to $\textcolor{red}{- 5}$.

Let's list the factors of -6:

$\textcolor{b l u e}{1 \times - 6 = - 6}$
$2 \times - 3 = - 6$
$3 \times - 2 = - 6$
$6 \times - 1 = - 6$

From the above combinations, $1 + \left(- 6\right) = 1 - 6 = - 5$.

Now back to the quadratic equation:

$2 {x}^{2} - 5 x - 3 = 0$

Here, we write $- 5$ as $1 - 6$.
Then $- 5 x$ will be $x - 6 x$

$\implies 2 {x}^{2} \textcolor{red}{+ 1 x - 6 x} - 3 = 0$

Make 2 pairs:
$\implies \textcolor{\mathmr{and} a n \ge}{2 {x}^{2} + x} \textcolor{b l u e}{- 6 x - 3} = 0$

Take out the common terms from each pair:

$\implies \textcolor{\mathmr{and} a n \ge}{x \left(2 x + 1\right)} \textcolor{b l u e}{- 3 \left(2 x + 1\right)} = 0$

$\textcolor{red}{2 x + 1}$ is common to the two terms in the equation. Take out the common terms, and write the remaining terms:

$\implies \textcolor{red}{\left(2 x + 1\right)} \textcolor{g r e e n}{\left(x - 3\right)} = 0$

$y = \left(2 x + 1\right) \left(x - 3\right)$

We can check our answer by working backwards:

i.e. solve: $y = \left(2 x + 1\right) \left(x - 3\right)$

$y = 2 x \left(x - 3\right) + 1 \left(x - 3\right)$

$y = 2 {x}^{2} - 6 x + x - 3$

$y = 2 {x}^{2} - 5 x - 3$