 Problem of the Week Updated at May 13, 2019 8:42 AM

How can we factor $$30{v}^{2}-27v+6$$?

Below is the solution.

$30{v}^{2}-27v+6$

 1 How?Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$30{v}^{2}$$, $$-27v$$, and $$6$$?It is $$3$$.2 What is the highest degree of $$v$$ that divides evenly into $$30{v}^{2}$$, $$-27v$$, and $$6$$?It is 1, since $$v$$ 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{30{v}^{2}}{3}+\frac{-27v}{3}+\frac{6}{3})$3 Simplify each term in parentheses.$3(10{v}^{2}-9v+2)$4 How?Split the second term in $$10{v}^{2}-9v+2$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$10\times 2=20$2 Ask: Which two numbers add up to $$-9$$ and multiply to $$20$$?$$-4$$ and $$-5$$3 Split $$-9v$$ as the sum of $$-4v$$ and $$-5v$$.$10{v}^{2}-4v-5v+2$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$3(10{v}^{2}-4v-5v+2)$5 Factor out common terms in the first two terms, then in the last two terms.$3(2v(5v-2)-(5v-2))$6 Factor out the common term $$5v-2$$.$3(5v-2)(2v-1)$Done 3*(5*v-2)*(2*v-1)