# Problem of the Week

## Updated at May 21, 2018 4:23 PM

This week we have another algebra problem:

How can we compute the factors of $$12{x}^{2}-27x+15$$?

Let's start!

$12{x}^{2}-27x+15$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$12{x}^{2}$$, $$-27x$$, and $$15$$?It is $$3$$.2 What is the highest degree of $$x$$ that divides evenly into $$12{x}^{2}$$, $$-27x$$, and $$15$$?It is 1, since $$x$$ 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{12{x}^{2}}{3}+\frac{-27x}{3}+\frac{15}{3})$3 Simplify each term in parentheses.$3(4{x}^{2}-9x+5)$4 Split the second term in $$4{x}^{2}-9x+5$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$4\times 5=20$2 Ask: Which two numbers add up to $$-9$$ and multiply to $$20$$?$$-4$$ and $$-5$$3 Split $$-9x$$ as the sum of $$-4x$$ and $$-5x$$.$4{x}^{2}-4x-5x+5$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$3(4{x}^{2}-4x-5x+5)$5 Factor out common terms in the first two terms, then in the last two terms.$3(4x(x-1)-5(x-1))$6 Factor out the common term $$x-1$$.$3(x-1)(4x-5)$Done 3*(x-1)*(4*x-5)