How do you factor completely $45x^3+93x^2+30x$ ?

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Answer 1 · 2015-11-19T10:51:47

Take out the common factor of 3x and then factor the quadratic as normal

$45x^3 +93x^2 +30x$ = $3x (15x^2 + 31x + 10)$
$3x(15x^2 +31x +10)$ = $3x(5x + 2)(3x + 5)$

How do you factor completely 45x^3+93x^2+30x45x^3+93x^2+30x?