# How do you factor y= 2x^2 - 5x -3 ?

Dec 14, 2015

Use the quadratic formula to find:

$y = 2 {x}^{2} - 5 x - 3 = \left(2 x + 1\right) \left(x - 3\right)$

#### Explanation:

This quadratic is of the form $a {x}^{2} + b x + c$ with $a = 2$, $b = - 5$ and $c = - 3$. It has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{5 \pm \sqrt{{5}^{2} - \left(4 \times 2 \times - 3\right)}}{2 \times 2}$

$= \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4}$

That is $x = - \frac{1}{2}$ and $x = 3$

Hence our quadratic has linear factors $\left(2 x + 1\right)$ and $\left(x - 3\right)$

$y = 2 {x}^{2} - 5 x - 3 = \left(2 x + 1\right) \left(x - 3\right)$

Dec 14, 2015

Find a suitable split for the middle term, then factor by grouping to find:

$y = 2 {x}^{2} - 5 x - 3 = \left(2 x + 1\right) \left(x - 3\right)$

#### Explanation:

Look for a pair of factors of $A C = 2 \cdot 3 = 6$ whose difference is $B = 5$

The pair $6 , 1$ works. Use that to split the middle term then factor by grouping as follows:

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

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

$= \left(2 {x}^{2} - 6 x\right) + \left(x - 3\right)$

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

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