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find a polynomial degree of 3 that has 3, -3, and 9
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multiply (x-3)(x+3)(x-9)=y
that has zeros 3,-3, and 9*
after you do (x-3)(x+3) you get x2+3x-3x-9. the 3's cancel out but then what is left just x or nothing at all?
The 3's cancel out, and you're then left with: x^2 - 9 Then you multiply that with (x - 9): (x^2 - 9)(x - 9) x^3 - 9x^2 - 9x + 81
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