How do you factor #5n^2+10n+20#?

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Answer 1 · 2015-05-19T00:11:30

y = 5(n^2 + 2n + 4)

The expression in parentheses can't be factored because D < 0

Answer 2 · 2015-05-19T00:19:42

We must find its roots and then transform them into factors.

Using Bhaskara we get:

#(-10+-sqrt(-300))/10#

However, #Delta=-300# indicates imaginary roots, because we can rewrite it as #300(-1)#, which will pave the way to find a solution.

We know, by imaginary numbers definition, that #(-1)=i^2#. Therefore, #sqrt(-1)=i# and we can proceed.

#(-10+-10isqrt(3))/10#=#-1+-isqrt(3)#

Imaginary roots cannot be factors.