Question

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 1

Answer:

$(4x-1)(4x+1)(16x^2+1)y^2$

False because C)

Step-by-step explanation:

Our original function is $256x^4 y^2-y^2$

The first step we go is to take $y^2$ as GCF and take it out of bracket

$y^2(256x^4-1)$

also $256=16^2$ and $x^4=(x^2)^2$ and $1=1^2$

Hence we can write the above equation as

$y^2((16x^2)^2-1^2)$

Now we apply the differences of the square in this which says

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

$((16x^2)^2-1^2)=(16x^2-1)(16x^2+1)$

$((16x^2)^2-1^2)=(4^2x^2-1^2)(16x^2+1)$

Again applying the differences of the square in the first factor we get

$(4x-1)(4x+1)(16x^2-+1)$

Hence

$y^2((16x^2)^2-1^2)=y^2(4x-1)(4x+1)(16x^2-+1)$

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.