How do you completely factor $x^{3} + 4 x^{2} - 45 x$ $x^3 + 4x^2 - 45x$ ?

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Answer 1 · 2018-06-12T21:54:02

$x (x - 5) (x + 9)$ $x(x-5)(x+9)$

take the $x$ $x$ as a common factor such that:

$x (x^{2} + 4 x - 45)$ $x(x^{2}+4x-45)$

factorize the expression inside the brackets

$x (x^{2} + 9 x - 5 x - 45)$ $x(x^{2} + 9x-5x-45)$

$x (x (x + 9) - 5 (x + 9))$ $x(x(x+9)-5(x+9))$
$x (x - 5) (x + 9)$ $x(x-5)(x+9)$

How do you completely factor x^3 + 4x^2 - 45xx3+4x2−45xx^3 + 4x^2 - 45x?