Is it possible to factor $y = x^{3} + 5 x^{2} - 14 x$ $y=x^3+5x^2-14x$ ? If so, what are the factors?

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Answer 1 · 2015-12-15T04:33:24

Yes; $x = - 7, 0, 2$ $x=-7, 0, 2$

Original Equation:
$y = x^{3} + 5 x^{2} - 14 x$ $y=x^3+5x^2-14x$

Factor out an $x$ $x$ :
$y = x (x^{2} + 5 x - 14)$ $y=x(x^2+5x-14)$

Factor inside the parentheses (factors of $A C$ $AC$ that add up to $B$ $B$ ):
––Factors of $- 14$ $-14$ that add up to $5$ $5$

You should get $7$ $7$ and $- 2$ $-2$

Now plug it back into the equation:
$y = x (x - 2) (x + 7)$ $y=x(x-2)(x+7)$

Your factors are:
$x$ $x$
$(x - 2)$ $(x-2)$
$(x + 7)$ $(x+7)$

To find $x$ $x$ , set each factor equal to $0$ $0$ and solve.
Your final answer should be:
$x = 0$ $x=0$
$x = 2$ $x=2$
$x = - 7$ $x=-7$

Is it possible to factor y=x3+5x2−14xy=x^3+5x^2-14x ? If so, what are the factors?