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

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Answer 1 · 2016-07-14T17:21:58

$y {(8 y + 5)}^{2}$ $y(8y+5)^2$

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

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

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

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

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

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

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

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

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

Can 64y3+80y2+25y64y^3 + 80y^2 + 25y be factored? If so what are the factors ?