# Problem of the Week

Updated at Dec 17, 2018 11:09 AM

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

How can we factor $$36{v}^{2}-48v+15$$?

Check out the solution below!

$36{v}^{2}-48v+15$

 1 How?Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$36{v}^{2}$$, $$-48v$$, and $$15$$?It is $$3$$.2 What is the highest degree of $$v$$ that divides evenly into $$36{v}^{2}$$, $$-48v$$, and $$15$$?It is 1, since $$v$$ 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{36{v}^{2}}{3}+\frac{-48v}{3}+\frac{15}{3})$3 Simplify each term in parentheses.$3(12{v}^{2}-16v+5)$4 How?Split the second term in $$12{v}^{2}-16v+5$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$12\times 5=60$2 Ask: Which two numbers add up to $$-16$$ and multiply to $$60$$?$$-6$$ and $$-10$$3 Split $$-16v$$ as the sum of $$-6v$$ and $$-10v$$.$12{v}^{2}-6v-10v+5$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$3(12{v}^{2}-6v-10v+5)$5 Factor out common terms in the first two terms, then in the last two terms.$3(6v(2v-1)-5(2v-1))$6 Factor out the common term $$2v-1$$.$3(2v-1)(6v-5)$Done3*(2*v-1)*(6*v-5)