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

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Answer 1 · 2015-02-05T10:52:08

Okay so the first step will be removing the gcf

So lets write this as $n (n^2 + 2n- 35)$
so lets faCtor $(n^2 + 2n- 35)$

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

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 $(0,-7,5)$

Answer 2 · 2015-02-05T11:01:04

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

$n^3+2n^2-35n=n*(n^2+2n-35)$

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

Let's look at what we have left. A quadratic equation of the form
$ax^2+bx+c=0$ 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 $(n+7)(n-5)$

And the whole original equation will factor into:

$n^3+2n^2-35n=n*(n+7)(n-5)=0$

Answer :
$n=0orn=-7orn=5$

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