Solve z^2+8z-4z-32

$$z ^ { 2 } + 8 z - 4 z - 32$$

Factor

$$\left(z-4\right)\left(z+8\right)$$

Solution Steps

Multiply and combine like terms.

$$z^{2}+4z-32$$

Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz-32. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-32\right)=-32$$

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 -32.

$$-1,32 -2,16 -4,8$$

Calculate the sum for each pair.

$$-1+32=31 -2+16=14 -4+8=4$$

The solution is the pair that gives sum 4.

$$a=-4 b=8$$

Rewrite z^{2}+4z-32 as \left(z^{2}-4z\right)+\left(8z-32\right).

$$\left(z^{2}-4z\right)+\left(8z-32\right)$$

Factor out z in the first and 8 in the second group.

$$z\left(z-4\right)+8\left(z-4\right)$$

Factor out common term z-4 by using distributive property.

$$\left(z-4\right)\left(z+8\right)$$

Evaluate

$$\left(z-4\right)\left(z+8\right)$$

Solution Steps

Combine 8z and -4z to get 4z.

$$z^{2}+4z-32$$