Question

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

Answer 1

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

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`