# How do you factor the expression 8y²-28y-60?

Jun 30, 2018

$4 \left(y - 5\right) \left(3 + 2 y\right)$

#### Explanation:

At first we Can factor $4 \left(2 {y}^{2} - 7 y - 15\right)$ since
$8 = 2 \cdot 4$
$28 = 7 \cdot 4$
and
$60 = 15 \cdot 4$

so we get

$4 \left(2 {y}^{2} - 7 y - 15\right)$
now we solve the equation

$2 {y}^{2} - 7 y - 15 = 0$
dividing by $2$
${y}^{2} - \frac{7}{2} y - \frac{15}{2} = 0$
by the quadratic formula we get

${y}_{1 , 2} = \frac{7}{4} \pm \sqrt{\frac{49}{16} + \frac{120}{16}}$
${y}_{1 , 2} = \frac{7}{4} \pm \sqrt{\frac{169}{16}}$
so we get

${y}_{1 , 2} = \frac{7}{4} \pm \frac{13}{4}$
this gives

${y}_{1} = 5$
${y}_{2} = \frac{6}{4} = \frac{3}{2}$

so we get

$4 \left(y - 5\right) \left(2 y + 3\right)$