# Problem of the Week

## Updated at Oct 17, 2022 11:23 AM

This week we have another algebra problem:

How can we compute the factors of $$12{n}^{2}+38n-14$$?

Let's start!

$12{n}^{2}+38n-14$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$12{n}^{2}$$, $$38n$$, and $$-14$$?It is $$2$$.2 What is the highest degree of $$n$$ that divides evenly into $$12{n}^{2}$$, $$38n$$, and $$-14$$?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{12{n}^{2}}{2}+\frac{38n}{2}-\frac{14}{2})$3 Simplify each term in parentheses.$2(6{n}^{2}+19n-7)$4 Split the second term in $$6{n}^{2}+19n-7$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$6\times -7=-42$2 Ask: Which two numbers add up to $$19$$ and multiply to $$-42$$?$$21$$ and $$-2$$3 Split $$19n$$ as the sum of $$21n$$ and $$-2n$$.$6{n}^{2}+21n-2n-7$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$2(6{n}^{2}+21n-2n-7)$5 Factor out common terms in the first two terms, then in the last two terms.$2(3n(2n+7)-(2n+7))$6 Factor out the common term $$2n+7$$.$2(2n+7)(3n-1)$Done2*(2*n+7)*(3*n-1)