How to factorize this #(16-x^2)(4-y)^2-(4-x)^2(16-y^2)#?

1 Answer
Jim G.
Jun 1, 2018

#8(4-x)(4-y)(x-y)#

Explanation:

#16-x^2" and "16-y^2" are "color(blue)"difference of squares"#

#•color(white)(x)a^2-b^2=(a-b)(a+b)#

#16-x^2=(4-x)(4+x)#

#16-y^2=(4-y)(4+y)#

#"we can now write the expression as"#

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

#"take out a "color(blue)"common factor "(4-x)(4-y)#

#=(4-x)(4-y)[(4+x)(4-y)-(4-x)(4+y)]#

#"expand the factors inside the square bracket"#

#(16-4y+4x-xy-(16+4y-4x-xy))#

#=(cancel(16)-4y+4xcancel(-xy)cancel(-16)-4y+4xcancel(+xy))#

#=(8x-8y)=8(x-y)#

#"putting it all together gives"#

#8(4-x)(4-y)(x-y)#

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