# Problem of the Week

## Updated at Feb 13, 2023 2:18 PM

This week we have another algebra problem:

How can we factor $$6{y}^{2}-48y+42$$?

Let's start!

$6{y}^{2}-48y+42$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$6{y}^{2}$$, $$-48y$$, and $$42$$?It is $$6$$.2 What is the highest degree of $$y$$ that divides evenly into $$6{y}^{2}$$, $$-48y$$, and $$42$$?It is 1, since $$y$$ 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{6{y}^{2}}{6}+\frac{-48y}{6}+\frac{42}{6})$3 Simplify each term in parentheses.$6({y}^{2}-8y+7)$4 Factor $${y}^{2}-8y+7$$.1 Ask: Which two numbers add up to $$-8$$ and multiply to $$7$$?$$-7$$ and $$-1$$2 Rewrite the expression using the above.$(y-7)(y-1)$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$6(y-7)(y-1)$Done 6*(y-7)*(y-1)