# Problem of the Week

## Updated at Oct 10, 2022 3:56 PM

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

How can we compute the factors of $$15{q}^{2}+6q-21$$?

Here are the steps:

$15{q}^{2}+6q-21$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$15{q}^{2}$$, $$6q$$, and $$-21$$?It is $$3$$.2 What is the highest degree of $$q$$ that divides evenly into $$15{q}^{2}$$, $$6q$$, and $$-21$$?It is 1, since $$q$$ 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{15{q}^{2}}{3}+\frac{6q}{3}-\frac{21}{3})$3 Simplify each term in parentheses.$3(5{q}^{2}+2q-7)$4 Split the second term in $$5{q}^{2}+2q-7$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$5\times -7=-35$2 Ask: Which two numbers add up to $$2$$ and multiply to $$-35$$?$$7$$ and $$-5$$3 Split $$2q$$ as the sum of $$7q$$ and $$-5q$$.$5{q}^{2}+7q-5q-7$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$3(5{q}^{2}+7q-5q-7)$5 Factor out common terms in the first two terms, then in the last two terms.$3(q(5q+7)-(5q+7))$6 Factor out the common term $$5q+7$$.$3(5q+7)(q-1)$Done3*(5*q+7)*(q-1)