Solve 6y^3+14y^2+4y

$$6 y ^ { 3 } + 14 y ^ { 2 } + 4 y$$

Factor

$$2y\left(y+2\right)\left(3y+1\right)$$

Solution Steps

Factor out 2.

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

Consider 3y^{3}+7y^{2}+2y. Factor out y.

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

Consider 3y^{2}+7y+2. Factor the expression by grouping. First, the expression needs to be rewritten as 3y^{2}+ay+by+2. To find a and b, set up a system to be solved.

$$a+b=7 ab=3\times 2=6$$

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 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 7.

$$a=1 b=6$$

Rewrite 3y^{2}+7y+2 as \left(3y^{2}+y\right)+\left(6y+2\right).

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

Factor out y in the first and 2 in the second group.

$$y\left(3y+1\right)+2\left(3y+1\right)$$

Factor out common term 3y+1 by using distributive property.

$$\left(3y+1\right)\left(y+2\right)$$

Rewrite the complete factored expression.

$$2y\left(3y+1\right)\left(y+2\right)$$

Evaluate

$$2y\left(y+2\right)\left(3y+1\right)$$