# How do you solve the equation n^3+2n^2-35n=0?

Feb 5, 2015

Okay so the first step will be removing the gcf

So lets write this as $n \left({n}^{2} + 2 n - 35\right)$
so lets faCtor$\left({n}^{2} + 2 n - 35\right)$

this CAN BE REWRITTEN AS $n \left(n + 7\right) \left(n - 5\right)$

Now using the 0 theorem we can say that one of the 3 polynomials which are being multiplied have to be = 0

So hence the set of values for n are $\left(0 , - 7 , 5\right)$

Feb 5, 2015

You can take one $n$ out and have a quadratic equation.

${n}^{3} + 2 {n}^{2} - 35 n = n \cdot \left({n}^{2} + 2 n - 35\right)$

Now we allready have one possible solution: $n = 0$

Let's look at what we have left. A quadratic equation of the form
$a {x}^{2} + b x + c = 0$ where $a = 1$, $b = 2$ and $c = - 35$

Now we have to find two numbers that, when multiplied, give a product of $- 35$ and when added/subtracted give a sum of $2$

These would be $7$ and $- 5$
So the quadratic part factors into $\left(n + 7\right) \left(n - 5\right)$

And the whole original equation will factor into:

${n}^{3} + 2 {n}^{2} - 35 n = n \cdot \left(n + 7\right) \left(n - 5\right) = 0$

$n = 0 \mathmr{and} n = - 7 \mathmr{and} n = 5$