# Problem of the Week

## Updated at Apr 8, 2024 1:34 PM

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

How would you find the factors of $$12{m}^{2}-18m+6$$?

Check out the solution below!

$12{m}^{2}-18m+6$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$12{m}^{2}$$, $$-18m$$, and $$6$$?It is $$6$$.2 What is the highest degree of $$m$$ that divides evenly into $$12{m}^{2}$$, $$-18m$$, and $$6$$?It is 1, since $$m$$ is not in every term.3 Multiplying the results above,The GCF is $$6$$.To get access to all 'How?' and 'Why?' steps, join Cymath Plus!GCF = $$6$$2 Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)$6(\frac{12{m}^{2}}{6}+\frac{-18m}{6}+\frac{6}{6})$3 Simplify each term in parentheses.$6(2{m}^{2}-3m+1)$4 Split the second term in $$2{m}^{2}-3m+1$$ 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 $$-3$$ and multiply to $$2$$?$$-1$$ and $$-2$$3 Split $$-3m$$ as the sum of $$-m$$ and $$-2m$$.$2{m}^{2}-m-2m+1$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$6(2{m}^{2}-m-2m+1)$5 Factor out common terms in the first two terms, then in the last two terms.$6(m(2m-1)-(2m-1))$6 Factor out the common term $$2m-1$$.$6(2m-1)(m-1)$Done6*(2*m-1)*(m-1)