Problem of the Week

Updated at Jan 27, 2014 4:04 PM

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Mar 31, 2025

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}

Find the Greatest Common Factor (GCF).

What is the largest number that divides evenly into

12{y}^{4}

100{y}^{3}

, and

112{y}^{2}

It is

4

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}

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}

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}})

Simplify each term in parentheses.

4{y}^{2}(3{y}^{2}+25y+28)

Split the second term in

3{y}^{2}+25y+28

into two terms.

Multiply the coefficient of the first term by the constant term.

3\times 28=84

Ask: Which two numbers add up to

25

and multiply to

84

21

and

4

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)

Factor out common terms in the first two terms, then in the last two terms.

4{y}^{2}(3y(y+7)+4(y+7))

Factor out the common term

y+7

4{y}^{2}(y+7)(3y+4)

Done