Solve -2y^3+3y^2+20y

$$- 2 y ^ { 3 } + 3 y ^ { 2 } + 20 y$$

Factor

$$y\left(4-y\right)\left(2y+5\right)$$

Solution Steps

Factor out y.

$$y\left(-2y^{2}+3y+20\right)$$

Consider -2y^{2}+3y+20. Factor the expression by grouping. First, the expression needs to be rewritten as -2y^{2}+ay+by+20. To find a and b, set up a system to be solved.

$$a+b=3 ab=-2\times 20=-40$$

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 -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 3.

$$a=8 b=-5$$

Rewrite -2y^{2}+3y+20 as \left(-2y^{2}+8y\right)+\left(-5y+20\right).

$$\left(-2y^{2}+8y\right)+\left(-5y+20\right)$$

Factor out 2y in the first and 5 in the second group.

$$2y\left(-y+4\right)+5\left(-y+4\right)$$

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

$$\left(-y+4\right)\left(2y+5\right)$$

Rewrite the complete factored expression.

$$y\left(-y+4\right)\left(2y+5\right)$$

Evaluate

$$y\left(4-y\right)\left(2y+5\right)$$