Question 970385
Let {{{Q = x^2 - 9}}}


{{{ (x^2-9)^2 + 8x (x^2-9) }}} will turn into {{{ Q^2 + 8x*Q }}}


Factor out the common term Q to get {{{ Q^2 + 8x*Q = Q*(Q + 8x) }}}


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{{{Q*(Q + 8x)}}}


{{{(x^2-9)*((x^2-9) + 8x)}}} replace every Q with x^2 - 9


{{{(x^2-9)*(x^2 + 8x-9)}}} Rearrange terms


{{{(x-3)(x+3)*(x^2 + 8x - 9)}}} Factor the first {{{x^2-9}}} (difference of squares rule)


{{{(x-3)(x+3)(x+9)(x-1)}}} Factor {{{x^2 + 8x - 9}}} (note: 9 and -1 multiply to -9 and add to +8)


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So in the end, {{{ (x^2-9)^2 + 8x (x^2-9) }}} factors to {{{(x-3)(x+3)(x+9)(x-1)}}}