# Can 64y^3 + 80y^2 + 25y  be factored? If so what are the factors ?

Jul 14, 2016

$y {\left(8 y + 5\right)}^{2}$

#### Explanation:

First we can take a common factor of $y$ outside:

$y \left(64 {y}^{2} + 80 y + 25\right)$

Leaving us with a quadratic inside the brackets which we can use the quadratic formula on:

$y = \frac{- 80 \pm \sqrt{{80}^{2} - 4 \left(64\right) \left(25\right)}}{128}$

Notice that the square root term is just zero, so we have a double root of $- \frac{80}{128} = - \frac{5}{8}$

This means we need a bracket that we can square which will give the original quadratic while also resulting in $y = - \frac{5}{8}$

Simplest to look at the squared term, $\sqrt{64 {y}^{2}} = 8 y$ so:

y(8y+color(red)(?))^2 = y(64y^2+80y+25)

In order for the left hand side to have roots of $y = 0 \mathmr{and} y = - \frac{5}{8}$ must have color(red)(?) = 5