How do you solve the equation n3+2n2−35n=0?

Question

jojoann325 · Accepted Answer

Step 1  So lets write this as  n(n2+2n−35)  So lets factor  (n2+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)

Joy Compton · Accepted Answer

Step 1
Can take one n out and have a quadratic equation.
$n^{3} + 2 n^{2} - 35 n = n \times (n^{2} + 2 n - 35)$
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 $(n + 7) (n - 5)$
And the whole original equation will factor into:
$n^{3} + 2 n^{2} - 35 n = n \times (n + 7) (n - 5) = 0$
Answer:
$n = 0 or n = - 7$ or $n = 5$

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

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