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anonymous
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\[x(x^29)\] is a start. then factor the \(x^29\) part as the difference of two squares

tpenn
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wait im confused, can you show work?

hawkfalcon
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He just took out an x

tpenn
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so x(x^29) is the answer?

hawkfalcon
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\[x^39x\]
x is common to both sides

hawkfalcon
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so then \[x(x^29)\]

tpenn
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ohh okay

hawkfalcon
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But \[x^29 is factorable\]

tpenn
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how do you factor x^29 though?

hawkfalcon
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Use the rule of a difference of two squares.

hawkfalcon
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\[(x+3)(x3)\]

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