Question 131978
{{{25x^2-40x+16}}}


Well, if each factor had a 5x and each factor had a -4, that would take care of the {{{x^2}}} and constant terms, so let's see what happens to the middle term:


{{{(5x-4)(5x-4)=25x^2-20x-20x+16=25x^2-40x+16}}}.   Yep, that does it.


{{{x^3+2x^2-3x}}}.  This one has at least one x in every term, so you can start by factoring out an x.


{{{x(x^2+2x-3)}}}.


Now, {{{3*-1=-3}}} and {{{3+(-1)=2}}}, so the rest of this thing should factor to:


{{{x(x-1)(x+3)}}}


Check by multiplying it out:  {{{x(x-1)(x+3)=x(x^2+3x-x-3)=x(x^2+2x-3)=x^3+2x^2-3x}}}.  Yep, that works.