Solve 7y^4+42y^3+56y^2

$$7 { y }^{ 4 } +42 { y }^{ 3 } +56 { y }^{ 2 }$$

Factor

$$7\left(y+2\right)\left(y+4\right)y^{2}$$

Solution Steps

Factor out 7.

$$7\left(y^{4}+6y^{3}+8y^{2}\right)$$

Consider y^{4}+6y^{3}+8y^{2}. Factor out y^{2}.

$$y^{2}\left(y^{2}+6y+8\right)$$

Consider y^{2}+6y+8. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+8. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\times 8=8$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=2 b=4$$

Rewrite y^{2}+6y+8 as \left(y^{2}+2y\right)+\left(4y+8\right).

$$\left(y^{2}+2y\right)+\left(4y+8\right)$$

Factor out y in the first and 4 in the second group.

$$y\left(y+2\right)+4\left(y+2\right)$$

Factor out common term y+2 by using distributive property.

$$\left(y+2\right)\left(y+4\right)$$

Rewrite the complete factored expression.

$$7y^{2}\left(y+2\right)\left(y+4\right)$$

Evaluate

$$7\left(y+2\right)\left(y+4\right)y^{2}$$