# Problem of the Week

## Updated at Mar 9, 2020 5:14 PM

To get more practice in algebra, we brought you this problem of the week:

How can we compute the factors of $$28{p}^{2}-14p-14$$?

Check out the solution below!

$28{p}^{2}-14p-14$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$28{p}^{2}$$, $$-14p$$, and $$-14$$?It is $$14$$.2 What is the highest degree of $$p$$ that divides evenly into $$28{p}^{2}$$, $$-14p$$, and $$-14$$?It is 1, since $$p$$ is not in every term.3 Multiplying the results above,The GCF is $$14$$.To get access to all 'How?' and 'Why?' steps, join Cymath Plus!GCF = $$14$$2 Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)$14(\frac{28{p}^{2}}{14}+\frac{-14p}{14}-\frac{14}{14})$3 Simplify each term in parentheses.$14(2{p}^{2}-p-1)$4 Split the second term in $$2{p}^{2}-p-1$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$2\times -1=-2$2 Ask: Which two numbers add up to $$-1$$ and multiply to $$-2$$?$$1$$ and $$-2$$3 Split $$-p$$ as the sum of $$p$$ and $$-2p$$.$2{p}^{2}+p-2p-1$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$14(2{p}^{2}+p-2p-1)$5 Factor out common terms in the first two terms, then in the last two terms.$14(p(2p+1)-(2p+1))$6 Factor out the common term $$2p+1$$.$14(2p+1)(p-1)$Done14*(2*p+1)*(p-1)