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.

How do you factor 5n^2+10n+205n^2+10n+20?