# Problem of the Week

## Updated at Aug 6, 2018 10:10 AM

This week's problem comes from the algebra category.

How would you find the factors of $$12{n}^{2}-10n-42$$?

Let's begin!

$12{n}^{2}-10n-42$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$12{n}^{2}$$, $$-10n$$, and $$-42$$?It is $$2$$.2 What is the highest degree of $$n$$ that divides evenly into $$12{n}^{2}$$, $$-10n$$, and $$-42$$?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{-10n}{2}-\frac{42}{2})$3 Simplify each term in parentheses.$2(6{n}^{2}-5n-21)$4 Split the second term in $$6{n}^{2}-5n-21$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$6\times -21=-126$2 Ask: Which two numbers add up to $$-5$$ and multiply to $$-126$$?$$9$$ and $$-14$$3 Split $$-5n$$ as the sum of $$9n$$ and $$-14n$$.$6{n}^{2}+9n-14n-21$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$2(6{n}^{2}+9n-14n-21)$5 Factor out common terms in the first two terms, then in the last two terms.$2(3n(2n+3)-7(2n+3))$6 Factor out the common term $$2n+3$$.$2(2n+3)(3n-7)$Done 2*(2*n+3)*(3*n-7)