Question 999791

compare {{{16x^2(x+4)}}} and {{{8(x+4)}}}: 

in {{{16x^2(x+4)}}} first term is {{{16x^2}}} and second term is {{{(x+4)}}}

in {{{8(x+4)}}} first term is {{{8}}} and second term is {{{(x+4)}}}

since second terms are equal, make first terms equal too

{{{16x^2}}} will be equal to {{{8}}} only if  we multiply it by {{{2x^2}}}:{{{8*2x^2=16x^2}}}

then, {{{16x^2(x+4)}}} can be written as a product of {{{8(x+4)}}} and {{{2x^2}}}

{{{16x^2(x+4)=8(x+4)(2x^2)}}}

so, answer is:

D){{{2x^2}}}