Solve 6y^3+3y^2-18y

$$6 y ^ { 3 } + 3 y ^ { 2 } - 18 y$$

Factor

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

Solution Steps

Factor out 3.

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

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

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

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

$$a+b=1 ab=2\left(-6\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite 2y^{2}+y-6 as \left(2y^{2}-3y\right)+\left(4y-6\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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