# Problem of the Week

## Updated at Jan 27, 2014 4:04 PM

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

How would you find the factors of $$12{y}^{4}+100{y}^{3}+112{y}^{2}$$?

Check out the solution below!

$12{y}^{4}+100{y}^{3}+112{y}^{2}$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$12{y}^{4}$$, $$100{y}^{3}$$, and $$112{y}^{2}$$?It is $$4$$.2 What is the highest degree of $$y$$ that divides evenly into $$12{y}^{4}$$, $$100{y}^{3}$$, and $$112{y}^{2}$$?It is $${y}^{2}$$.3 Multiplying the results above,The GCF is $$4{y}^{2}$$.To get access to all 'How?' and 'Why?' steps, join Cymath Plus!GCF = $$4{y}^{2}$$2 Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)$4{y}^{2}(\frac{12{y}^{4}}{4{y}^{2}}+\frac{100{y}^{3}}{4{y}^{2}}+\frac{112{y}^{2}}{4{y}^{2}})$3 Simplify each term in parentheses.$4{y}^{2}(3{y}^{2}+25y+28)$4 Split the second term in $$3{y}^{2}+25y+28$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$3\times 28=84$2 Ask: Which two numbers add up to $$25$$ and multiply to $$84$$?$$21$$ and $$4$$3 Split $$25y$$ as the sum of $$21y$$ and $$4y$$.$3{y}^{2}+21y+4y+28$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$4{y}^{2}(3{y}^{2}+21y+4y+28)$5 Factor out common terms in the first two terms, then in the last two terms.$4{y}^{2}(3y(y+7)+4(y+7))$6 Factor out the common term $$y+7$$.$4{y}^{2}(y+7)(3y+4)$Done4*y^2*(y+7)*(3*y+4)