Solve 8y^3+6y^2-9y=

$$8 y ^ { 3 } + 6 y ^ { 2 } - 9 y =$$

Factor

$$y\left(4y-3\right)\left(2y+3\right)$$

Solution Steps

Factor out y.

$$y\left(8y^{2}+6y-9\right)$$

Consider 8y^{2}+6y-9. Factor the expression by grouping. First, the expression needs to be rewritten as 8y^{2}+ay+by-9. To find a and b, set up a system to be solved.

$$a+b=6 ab=8\left(-9\right)=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 6.

$$a=-6 b=12$$

Rewrite 8y^{2}+6y-9 as \left(8y^{2}-6y\right)+\left(12y-9\right).

$$\left(8y^{2}-6y\right)+\left(12y-9\right)$$

Factor out 2y in the first and 3 in the second group.

$$2y\left(4y-3\right)+3\left(4y-3\right)$$

Factor out common term 4y-3 by using distributive property.

$$\left(4y-3\right)\left(2y+3\right)$$

Rewrite the complete factored expression.

$$y\left(4y-3\right)\left(2y+3\right)$$

Evaluate

$$y\left(4y-3\right)\left(2y+3\right)$$