Solve -8y^2z-40yz+48z

$$- 8 y ^ { 2 } z - 40 y z + 48 z$$

Factor

$$8z\left(1-y\right)\left(y+6\right)$$

Solution Steps

Factor out 8.

$$8\left(-y^{2}z-5yz+6z\right)$$

Consider -y^{2}z-5yz+6z. Factor out z.

$$z\left(-y^{2}-5y+6\right)$$

Consider -y^{2}-5y+6. Factor the expression by grouping. First, the expression needs to be rewritten as -y^{2}+ay+by+6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-6=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=1 b=-6$$

Rewrite -y^{2}-5y+6 as \left(-y^{2}+y\right)+\left(-6y+6\right).

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

Factor out y in the first and 6 in the second group.

$$y\left(-y+1\right)+6\left(-y+1\right)$$

Factor out common term -y+1 by using distributive property.

$$\left(-y+1\right)\left(y+6\right)$$

Rewrite the complete factored expression.

$$8z\left(-y+1\right)\left(y+6\right)$$

Evaluate

$$8z\left(1-y\right)\left(y+6\right)$$