Question 99741
Factor completely using integer coefficients:
e) {{{x^4+8x^3-2x^2-16x}}} First, factor an x.
{{{x(x^3+8x^2-2x-16)}}} Now factor the parentheses. I know that my factors here will be two binomials of the form:
{{{(x^2+m)(x+n)}}} where {{{m*n = -16}}} and {{{nx^2 = 8x^2}}} and {{{mx = -2x}}}, so m = -2 and n = 8. So the parentheses, when factored, look like:
{{{(x^2-2)(x+8)}}}, then the final answer is:
{{{x^4+8x^3-2x^2-16x = x(x^2-2)(x+8)}}}

f) {{{2x^6-20x^4-16x^3+160x}}} First, factor 2x.
{{{2x(x^5-10x^3-8x^2+80)}}} Now factor the parentheses. I know my factors here will be two binomials of the form:{{{(x^3+m)(x^2+n)}}} where: {{{m*n = 80}}}{{{nx^3 = -10x^3}}} and {{{mx^2 = -8x^2}}} so {{{m = -8}}} and {{{n = -10}}}, so the parentheses, when factored, look like:
{{{x^3-8)(x^2-10)}}}, so far we have:
{{{2x^6-20x^4-16x^3+160x = 2x(x^3-8)(x^2-10)}}} but notice that {{{x^3-8}}} is the difference of two cubes and this can be factored.
{{{x^3-8 = (x)^3-(2)^3}}} The difference of two cubes is factored thus:
{{{A^3-B^3 = (A-B)(A^2+AB+B^2)}}} so, in this case:
{{{x^3-8 = (x-2)(x^2+2x+4)}}} putting it all together, we have:
{{{2x^6-20x^4-16x^3+160x = 2x(x-2)(x^2+2x+4)(x^2-10)}}}

j) {{{x^4-x^3-x+1}}} Factor as:
{{{(x^3-1)(x-1)}}} But again, we have the difference of two cubes:{{{(x)^3-(1)^3}}}, so...
{{{x^4-x^3-x+1 = (x-1)(x^2+x+1)(x-1)}}} or:
{{{x^4-x^3-x+1 = (x-1)^2(x^2+x+1)}}}