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\[x(x^2-9)\] is a start. then factor the \(x^2-9\) part as the difference of two squares

wait im confused, can you show work?

He just took out an x

so x(x^2-9) is the answer?

\[x^3-9x\]
x is common to both sides

so then \[x(x^2-9)\]

ohh okay

But \[x^2-9 is factorable\]

how do you factor x^2-9 though?

Use the rule of a difference of two squares.

\[(x+3)(x-3)\]

oh okay i understand, i was just confused at how you were explaining it, thanks