Question

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

Answer 1

`n(n-3)(n+7)`

Explanation:

Given: `n^3+4n^2-21n=0`.

Factor out `n` first.

`n(n^2+4n-21)=0`.

Factor `(n^2+4n-21)`.

`=>n(n^2+7n-3n-21)`

`=>n(n(n+7)-3(n+7))`

`=>n(n-3)(n+7)`

Answer 2

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

Explanation:

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

`n(n^2+4x-21)=0`

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

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