# Problem of the Week

## Updated at Mar 29, 2021 9:12 AM

How can we factor $$36{n}^{2}+6n-12$$?

Below is the solution.

$36{n}^{2}+6n-12$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$36{n}^{2}$$, $$6n$$, and $$-12$$?It is $$6$$.2 What is the highest degree of $$n$$ that divides evenly into $$36{n}^{2}$$, $$6n$$, and $$-12$$?It is 1, since $$n$$ is not in every term.3 Multiplying the results above,The GCF is $$6$$.To get access to all 'How?' and 'Why?' steps, join Cymath Plus!GCF = $$6$$2 Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)$6(\frac{36{n}^{2}}{6}+\frac{6n}{6}-\frac{12}{6})$3 Simplify each term in parentheses.$6(6{n}^{2}+n-2)$4 Split the second term in $$6{n}^{2}+n-2$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$6\times -2=-12$2 Ask: Which two numbers add up to $$1$$ and multiply to $$-12$$?$$4$$ and $$-3$$3 Split $$n$$ as the sum of $$4n$$ and $$-3n$$.$6{n}^{2}+4n-3n-2$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$6(6{n}^{2}+4n-3n-2)$5 Factor out common terms in the first two terms, then in the last two terms.$6(2n(3n+2)-(3n+2))$6 Factor out the common term $$3n+2$$.$6(3n+2)(2n-1)$Done 6*(3*n+2)*(2*n-1)