Solve 40-5y+8y-y^2

$$40 - 5 y + 8 y - y ^ { 2 }$$

Factor

$$-\left(y-8\right)\left(y+5\right)$$

Solution Steps

Multiply and combine like terms.

$$-y^{2}+3y+40$$

Factor the expression by grouping. First, the expression needs to be rewritten as -y^{2}+ay+by+40. To find a and b, set up a system to be solved.

$$a+b=3 ab=-40=-40$$

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 -y^{2}+3y+40 as \left(-y^{2}+8y\right)+\left(-5y+40\right).

$$\left(-y^{2}+8y\right)+\left(-5y+40\right)$$

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

$$-y\left(y-8\right)-5\left(y-8\right)$$

Factor out common term y-8 by using distributive property.

$$\left(y-8\right)\left(-y-5\right)$$

Evaluate

$$-\left(y-8\right)\left(y+5\right)$$

Solution Steps

Combine -5y and 8y to get 3y.

$$40+3y-y^{2}$$