# Problem of the Week

## Updated at May 30, 2022 9:40 AM

This week's problem comes from the algebra category.

How would you find the factors of $$20{z}^{2}+6z-2$$?

Let's begin!

$20{z}^{2}+6z-2$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$20{z}^{2}$$, $$6z$$, and $$-2$$?It is $$2$$.2 What is the highest degree of $$z$$ that divides evenly into $$20{z}^{2}$$, $$6z$$, and $$-2$$?It is 1, since $$z$$ 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{20{z}^{2}}{2}+\frac{6z}{2}-\frac{2}{2})$3 Simplify each term in parentheses.$2(10{z}^{2}+3z-1)$4 Split the second term in $$10{z}^{2}+3z-1$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$10\times -1=-10$2 Ask: Which two numbers add up to $$3$$ and multiply to $$-10$$?$$5$$ and $$-2$$3 Split $$3z$$ as the sum of $$5z$$ and $$-2z$$.$10{z}^{2}+5z-2z-1$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$2(10{z}^{2}+5z-2z-1)$5 Factor out common terms in the first two terms, then in the last two terms.$2(5z(2z+1)-(2z+1))$6 Factor out the common term $$2z+1$$.$2(2z+1)(5z-1)$Done2*(2*z+1)*(5*z-1)