How do you factor $p (x) = 2 x^{3} - x^{2} - 18 x + 9$ $p(x)= 2x^3-x^2-18x+9$ ?

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Answer 1 · 2015-09-19T03:07:10

$(x - 3) (x + 3) (2 x - 1)$ $(x-3)(x+3)(2x-1)$

$p (x) = 2 x^{3} - x^{2} - 18 x + 9$ $p(x)=2x^3−x^2−18x+9$

$= x^{2} (2 x - 1) - 9 (2 x - 1)$ $=x^2(2x - 1) -9(2x-1)$

$= (x^{2} - 9) (2 x - 1)$ $=(x^2-9)(2x-1)$

$= (x - 3) (x + 3) (2 x - 1)$ $=(x-3)(x+3)(2x-1)$

How do you factor p(x)= 2x^3-x^2-18x+9p(x)=2x3−x2−18x+9p(x)= 2x^3-x^2-18x+9?