Solve 3y^3+23y^2+14y

$$3 y ^ { 3 } + 23 y ^ { 2 } + 14 y$$

Factor

$$y\left(y+7\right)\left(3y+2\right)$$

Solution Steps

Factor out y.

$$y\left(3y^{2}+23y+14\right)$$

Consider 3y^{2}+23y+14. Factor the expression by grouping. First, the expression needs to be rewritten as 3y^{2}+ay+by+14. To find a and b, set up a system to be solved.

$$a+b=23 ab=3\times 14=42$$

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 42.

$$1,42 2,21 3,14 6,7$$

Calculate the sum for each pair.

$$1+42=43 2+21=23 3+14=17 6+7=13$$

The solution is the pair that gives sum 23.

$$a=2 b=21$$

Rewrite 3y^{2}+23y+14 as \left(3y^{2}+2y\right)+\left(21y+14\right).

$$\left(3y^{2}+2y\right)+\left(21y+14\right)$$

Factor out y in the first and 7 in the second group.

$$y\left(3y+2\right)+7\left(3y+2\right)$$

Factor out common term 3y+2 by using distributive property.

$$\left(3y+2\right)\left(y+7\right)$$

Rewrite the complete factored expression.

$$y\left(3y+2\right)\left(y+7\right)$$

Evaluate

$$y\left(y+7\right)\left(3y+2\right)$$