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

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Answer 1 · 2016-07-14T17:21:58

$y(8y+5)^2$

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

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

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

$y = (-80+-sqrt(80^2-4(64)(25)))/128$

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

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

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

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

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

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