# Problem of the Week

## Updated at Dec 30, 2019 1:33 PM

This week we have another algebra problem:

How can we factor $$4{w}^{2}-6w-10$$?

Let's start!

$4{w}^{2}-6w-10$

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