How do you factor $n^{3} + 4 n^{2} - 21 n = 0$ $n^3 + 4n^2 - 21n = 0$ ?

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How do you factor $n^{3} + 4 n^{2} - 21 n = 0$ $n^3 + 4n^2 - 21n = 0$ ?

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Answer 1 · 2018-05-10T13:36:35

$n (n - 3) (n + 7)$ $n(n-3)(n+7)$

Explanation:

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

Factor out $n$ $n$ first.

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

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

$\Rightarrow n (n^{2} + 7 n - 3 n - 21)$ $=>n(n^2+7n-3n-21)$

$\Rightarrow n (n (n + 7) - 3 (n + 7))$ $=>n(n(n+7)-3(n+7))$

$\Rightarrow n (n - 3) (n + 7)$ $=>n(n-3)(n+7)$

Answer 2 · 2018-05-10T13:39:01

$n (n - 3) (n + 7) = 0$ $n(n-3)(n+7)=0$

Explanation:

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

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

Notice that $3 \times 7 = 21$ $3xx7=21$ and that $7 - 3 = 4$ $7-3=4$ So we can use this in the form of:

$n (n - 3) (n + 7) = 0 d \leftarrow$ $n(n-3)(n+7)=0 color(white)("d")larr$ 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.

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

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How do you factor $n^{3} + 4 n^{2} - 21 n = 0$ $n^3 + 4n^2 - 21n = 0$ ?

2 Answers

Explanation:

Explanation:

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How do you factor n^3 + 4n^2 - 21n = 0n3+4n2−21n=0n^3 + 4n^2 - 21n = 0?

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How do you factor $n^{3} + 4 n^{2} - 21 n = 0$ $n^3 + 4n^2 - 21n = 0$ ?