# Problem of the Week

## Updated at Aug 27, 2018 12:06 PM

For this week we've brought you this algebra problem.

How can we factor $$7{n}^{2}-31n-20$$?

Here are the steps:

$7{n}^{2}-31n-20$

 1 Split the second term in $$7{n}^{2}-31n-20$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$7\times -20=-140$2 Ask: Which two numbers add up to $$-31$$ and multiply to $$-140$$?$$4$$ and $$-35$$3 Split $$-31n$$ as the sum of $$4n$$ and $$-35n$$.$7{n}^{2}+4n-35n-20$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$7{n}^{2}+4n-35n-20$2 Factor out common terms in the first two terms, then in the last two terms.$n(7n+4)-5(7n+4)$3 Factor out the common term $$7n+4$$.$(7n+4)(n-5)$Done(7*n+4)*(n-5)