# Problem of the Week

## Updated at Oct 24, 2022 12:52 PM

How would you find the factors of $$12{n}^{2}-36n-21$$?

Below is the solution.

$12{n}^{2}-36n-21$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$12{n}^{2}$$, $$-36n$$, and $$-21$$?It is $$3$$.2 What is the highest degree of $$n$$ that divides evenly into $$12{n}^{2}$$, $$-36n$$, and $$-21$$?It is 1, since $$n$$ is not in every term.3 Multiplying the results above,The GCF is $$3$$.To get access to all 'How?' and 'Why?' steps, join Cymath Plus!GCF = $$3$$2 Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)$3(\frac{12{n}^{2}}{3}+\frac{-36n}{3}-\frac{21}{3})$3 Simplify each term in parentheses.$3(4{n}^{2}-12n-7)$4 Split the second term in $$4{n}^{2}-12n-7$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$4\times -7=-28$2 Ask: Which two numbers add up to $$-12$$ and multiply to $$-28$$?$$2$$ and $$-14$$3 Split $$-12n$$ as the sum of $$2n$$ and $$-14n$$.$4{n}^{2}+2n-14n-7$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$3(4{n}^{2}+2n-14n-7)$5 Factor out common terms in the first two terms, then in the last two terms.$3(2n(2n+1)-7(2n+1))$6 Factor out the common term $$2n+1$$.$3(2n+1)(2n-7)$Done3*(2*n+1)*(2*n-7)