PLEASE HELP!!! WILL MARK BRAINLIEST!!!Spencer is asked to factor the polynomial 256x^4 y^2-y^2completely over the integers. His work is shown below.256x^(4)y^(2)=y^(2)(256x^(4)-1)y^(2)(256x^(4)-1)=y^2(16x^(2)+1)Did Spencer factor the polynomial completely over the integers? Why or why not?A. Spencer did factor the polynomial completely; he identified teh GCF and applied the difference of cubes method.B. Spencer did not factor the polynomial completely; 16x^(2)+1 can eb factored over the integers. C. Spencer did not factor the polynomial completely; 16x^(2)-1 can be factored over the integers. D. Spencer did factor the polynomial completely; he identified teh GCF and applied the difference of squares method.

Answer:[tex](4x-1)(4x+1)(16x^2+1)y^2[/tex]False because C)Step-by-step explanation:Our original function is [tex]256x^4 y^2-y^2[/tex]The first step we go is to take [tex]y^2[/tex] as GCF and take it out of bracket[tex]y^2(256x^4-1)[/tex]also [tex]256=16^2[/tex] and [tex]x^4=(x^2)^2[/tex] and [tex]1=1^2[/tex]Hence we can write the above equation as [tex]y^2((16x^2)^2-1^2)[/tex]Now we apply the differences of the square in this which says[tex]a^2-b^2= (a+b)(a-b)[/tex][tex]((16x^2)^2-1^2)=(16x^2-1)(16x^2+1)[/tex][tex]((16x^2)^2-1^2)=(4^2x^2-1^2)(16x^2+1)[/tex]Again applying the differences of the square in the first factor we get[tex](4x-1)(4x+1)(16x^2-+1)[/tex]Hence [tex]y^2((16x^2)^2-1^2)=y^2(4x-1)(4x+1)(16x^2-+1)[/tex]Hence Spencer did not factor the polynomial completely over the integers because Spencer did not factor the polynomial completely; 16x^(2)-1 can be factored over the integers.