# Problem of the Week

## Updated at Nov 15, 2021 12:21 PM

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

How can we factor $$18{n}^{2}-6n-4$$?

Here are the steps:

$18{n}^{2}-6n-4$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$18{n}^{2}$$, $$-6n$$, and $$-4$$?It is $$2$$.2 What is the highest degree of $$n$$ that divides evenly into $$18{n}^{2}$$, $$-6n$$, and $$-4$$?It is 1, since $$n$$ is not in every term.3 Multiplying the results above,The GCF is $$2$$.To get access to all 'How?' and 'Why?' steps, join Cymath Plus!GCF = $$2$$2 Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)$2(\frac{18{n}^{2}}{2}+\frac{-6n}{2}-\frac{4}{2})$3 Simplify each term in parentheses.$2(9{n}^{2}-3n-2)$4 Split the second term in $$9{n}^{2}-3n-2$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$9\times -2=-18$2 Ask: Which two numbers add up to $$-3$$ and multiply to $$-18$$?$$3$$ and $$-6$$3 Split $$-3n$$ as the sum of $$3n$$ and $$-6n$$.$9{n}^{2}+3n-6n-2$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$2(9{n}^{2}+3n-6n-2)$5 Factor out common terms in the first two terms, then in the last two terms.$2(3n(3n+1)-2(3n+1))$6 Factor out the common term $$3n+1$$.$2(3n+1)(3n-2)$Done 2*(3*n+1)*(3*n-2)