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#