# How do you factor n^3 + 4n^2 - 21n = 0?

May 10, 2018

$n \left(n - 3\right) \left(n + 7\right)$

#### Explanation:

Given: ${n}^{3} + 4 {n}^{2} - 21 n = 0$.

Factor out $n$ first.

$n \left({n}^{2} + 4 n - 21\right) = 0$.

Factor $\left({n}^{2} + 4 n - 21\right)$.

$\implies n \left({n}^{2} + 7 n - 3 n - 21\right)$

$\implies n \left(n \left(n + 7\right) - 3 \left(n + 7\right)\right)$

$\implies n \left(n - 3\right) \left(n + 7\right)$

May 10, 2018

$n \left(n - 3\right) \left(n + 7\right) = 0$

#### Explanation:

Notice that on the LHS we have $n$ in each term. So we can factor that out leaving a quadratic.

$n \left({n}^{2} + 4 x - 21\right) = 0$

Notice that $3 \times 7 = 21$ and that $7 - 3 = 4$ So we can use this in the form of:

$n \left(n - 3\right) \left(n + 7\right) = 0 \textcolor{w h i t e}{\text{d}} \leftarrow$ this is now factorised as required.
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The very fact that this is presented as an equation implies that you are required to determine feasible values for n that satisfy the given condition.

n=0;color(white)("..d") n=+3;color(white)("..d") n=-7