How do you factor completely $n^2 + 7n - 44$ ?

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Answer 1 · 2016-06-26T02:48:03

Find two numbers that multiply to $-44$ and that add to $7$ .

These two numbers are $11$ and $-4$ .

$n^2 + 7n - 44 = (n + 11)(n - 4)$

Hopefully this helps!

Answer 2 · 2016-06-26T02:57:56

Notice that $7=11-4$ hence

$n^2+7n-44=n^2+(11-4)n-44=n^2-4n+11n-44= n(n-4)+11(n-4)=(n+11)(n-4)$

How do you factor completely n^2 + 7n - 44n^2 + 7n - 44?