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