Question 281596
{{{x^3-4x}}}
for this one when u look at it theres a x in both variables so the gcf is x
{{{x(x^2-4)}}} whats left is a perfect square inside the brackets so you can factor further, i sugest getting familiar with what squars look like to make it easier for u to factor, otherwise u will need to use the quadratic formula.(Ex: {{{x^2-36}}}6's,{{{x^2-49}}}7's,{{{x^2-144}}}12's)
{{{highlight(x(x-2)(x+2))}}} this would be completly factored
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{{{(xy-x)+(8y-8)}}}use group factoring
{{{x(y-1)+8(y-1)}}}factor x and 8 out
{{{highlight((x+8)(y-1))}}}add them together, completely factored
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{{{x^3+x^2-90x}}} same as first one GCF=x
{{{x(x^2+x-90)}}} find 2 numbers multiply to get -90 and add to get 1. -9 and +10. if u cant figure how to do this then use quadratic formula.
{{{highlight(x(x+10)(x-9))}}}this is completely factored