How do you completely factor $x^{3} + 9 x^{2} + 12 x + 15$ $x^3+9x^2+12x+15$ given x+5 is one of the factors?

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Answer 1 · 2016-09-27T15:08:10

Let, $p (x) = x^{3} + 9 x^{2} + 12 x + 15$ $p(x)=x^3+9x^2+12x+15$ , and, we are given that $(x + 5)$ $(x+5)$

is a factor of $p (x)$ $p(x)$ .

But, this is not true , because,

$p (- 5) = - 125 + 225 - 60 + 15 = 55 \neq 0$ $p(-5)=-125+225-60+15=55!=0$

Let, $p (x) = x^{3} + 9 x^{2} + 12 x + 15$ $p(x)=x^3+9x^2+12x+15$ , and, we are given that $(x + 5)$ $(x+5)$

is a factor of $p (x)$ $p(x)$ .

But, this is not true , because,

$p (- 5) = - 125 + 225 - 60 + 15 = 55 \neq 0$ $p(-5)=-125+225-60+15=55!=0$

How do you completely factor x3+9x2+12x+15x^3+9x^2+12x+15 given x+5 is one of the factors?