# Problem of the Week

## Updated at Sep 12, 2022 11:29 AM

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

How can we compute the factors of $$8{y}^{2}-34y+30$$?

Check out the solution below!

$8{y}^{2}-34y+30$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$8{y}^{2}$$, $$-34y$$, and $$30$$?It is $$2$$.2 What is the highest degree of $$y$$ that divides evenly into $$8{y}^{2}$$, $$-34y$$, and $$30$$?It is 1, since $$y$$ 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{y}^{2}}{2}+\frac{-34y}{2}+\frac{30}{2})$3 Simplify each term in parentheses.$2(4{y}^{2}-17y+15)$4 Split the second term in $$4{y}^{2}-17y+15$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$4\times 15=60$2 Ask: Which two numbers add up to $$-17$$ and multiply to $$60$$?$$-5$$ and $$-12$$3 Split $$-17y$$ as the sum of $$-5y$$ and $$-12y$$.$4{y}^{2}-5y-12y+15$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$2(4{y}^{2}-5y-12y+15)$5 Factor out common terms in the first two terms, then in the last two terms.$2(y(4y-5)-3(4y-5))$6 Factor out the common term $$4y-5$$.$2(4y-5)(y-3)$Done2*(4*y-5)*(y-3)