How do you factor #4x ^ { 3} y ^ { 2} + x ^ { 2} y ^ { 2} - 28x y ^ { 2} - 7y ^ { 2}#?

2 Answers
Nallasivam V
Sep 23, 2017

#(x^2y^2-7y^2)(4x+1)#

Explanation:

Given -

#4x^3y^2+x^2y^2-28xy^2-7y^2#

#x^2y^2(4x+1)-7y^2(4x+1)#

#(x^2y^2-7y^2)(4x+1)#

George C.
Sep 23, 2017

#4x^3y^2+x^2y^2-28xy^2-7y^2#

#= y^2(x^2-7)(4x+1)#

#= y^2(x-sqrt(7))(x+sqrt(7))(4x+1)#

Explanation:

Given:

#4x^3y^2+x^2y^2-28xy^2-7y^2#

Note that all of the terms are divisible by #y^2#, so we can separate that out as a common factor.

Also note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this quadrinomial will factor by grouping.

We will also use the difference of squares identity, which can be written:

#a^2-b^2 = (a-b)(a+b)#

So:

#4x^3y^2+x^2y^2-28xy^2-7y^2 = y^2(4x^3+x^2-28x-7)#

#color(white)(4x^3y^2+x^2y^2-28xy^2-7y^2) = y^2((4x^3+x^2)-(28x+7))#

#color(white)(4x^3y^2+x^2y^2-28xy^2-7y^2) = y^2(x^2(4x+1)-7(4x+1))#

#color(white)(4x^3y^2+x^2y^2-28xy^2-7y^2) = y^2(x^2-7)(4x+1)#

#color(white)(4x^3y^2+x^2y^2-28xy^2-7y^2) = y^2(x^2-(sqrt(7))^2)(4x+1)#

#color(white)(4x^3y^2+x^2y^2-28xy^2-7y^2) = y^2(x-sqrt(7))(x+sqrt(7))(4x+1)#

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