# Problem of the Week

## Updated at Dec 3, 2018 9:51 AM

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

How can we factor $$8{p}^{2}-26p+6$$?

Here are the steps:

$8{p}^{2}-26p+6$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$8{p}^{2}$$, $$-26p$$, and $$6$$?It is $$2$$.2 What is the highest degree of $$p$$ that divides evenly into $$8{p}^{2}$$, $$-26p$$, and $$6$$?It is 1, since $$p$$ 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{8{p}^{2}}{2}+\frac{-26p}{2}+\frac{6}{2})$3 Simplify each term in parentheses.$2(4{p}^{2}-13p+3)$4 Split the second term in $$4{p}^{2}-13p+3$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$4\times 3=12$2 Ask: Which two numbers add up to $$-13$$ and multiply to $$12$$?$$-1$$ and $$-12$$3 Split $$-13p$$ as the sum of $$-p$$ and $$-12p$$.$4{p}^{2}-p-12p+3$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$2(4{p}^{2}-p-12p+3)$5 Factor out common terms in the first two terms, then in the last two terms.$2(p(4p-1)-3(4p-1))$6 Factor out the common term $$4p-1$$.$2(4p-1)(p-3)$Done 2*(4*p-1)*(p-3)