How do you factor $x^{4} - 2 x^{3} - 12 x^{2} + 18 x + 27$ $x^4-2x^3-12x^2+18x+27$ ?

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How do you factor $x^{4} - 2 x^{3} - 12 x^{2} + 18 x + 27$ $x^4-2x^3-12x^2+18x+27$ ?

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Answer 1 · 2015-05-07T16:46:19

Answer $(x + 1) {(x - 3)}^{2} (x + 3)$ $(x+1)(x-3)^2(x+3)$

Try $x = - 1$ $x=-1$ gives $1 + 2 - 12 - 18 + 27 = 0$ $1+2-12-18+27=0$ so $x + 1$ $x+1$ is a factor.
Dividing by $(x + 1)$ $(x+1)$ gives $x^{3} - 3 x^{2} - 9 x + 27$ $x^3-3x^2-9x+27$
Try $x = 3$ $x=3$ in cubic gives $27 - 27 - 27 + 27 = 0$ $27-27-27+27=0$ so $x - 3$ $x-3$ is a factor
Dividing by $(x - 3)$ $(x-3)$ gives $x^{2} - 9$ $x^2-9$ which factors to $(x - 3) (x + 3)$ $(x-3)(x+3)$

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How do you factor $x^{4} - 2 x^{3} - 12 x^{2} + 18 x + 27$ $x^4-2x^3-12x^2+18x+27$ ?

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How do you factor $x^{4} - 2 x^{3} - 12 x^{2} + 18 x + 27$ $x^4-2x^3-12x^2+18x+27$ ?