How do you factor $2 x^{3} - x^{2} - 162 x + 81$ $2x^3-x^2-162x+81$ ?

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Answer 1 · 2017-02-10T13:49:33

$= (2 x - 1) (x - 9) (x + 9)$ $=(2x-1)(x-9)(x+9)$

You could factor the bynomial

$2 x^{3} - x^{2} = x^{2} (2 x - 1)$ $2x^3-x^2=x^2(2x-1)$

and the bynomial

$- 162 x + 81 = - 81 (2 x - 1)$ $-162x+81=-81(2x-1)$

then the polynomial is:

$2 x^{3} - x^{2} - 162 x + 81 = x^{2} (2 x - 1) - 81 (2 x - 1)$ $2x^3-x^2-162x+81=x^2(2x-1)-81(2x-1)$

It's possible factor again:

$= (2 x - 1) (x^{2} - 81) = (2 x - 1) (x - 9) (x + 9)$ $=(2x-1)(x^2-81)=(2x-1)(x-9)(x+9)$

How do you factor 2x^3-x^2-162x+812x3−x2−162x+812x^3-x^2-162x+81?