Solve z^2+12z-3z-36

$$z ^ { 2 } + 12 z - 3 z - 36$$

Factor

$$\left(z-3\right)\left(z+12\right)$$

Solution Steps

Multiply and combine like terms.

$$z^{2}+9z-36$$

Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz-36. To find a and b, set up a system to be solved.

$$a+b=9 ab=1\left(-36\right)=-36$$

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 -36.

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

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 9.

$$a=-3 b=12$$

Rewrite z^{2}+9z-36 as \left(z^{2}-3z\right)+\left(12z-36\right).

$$\left(z^{2}-3z\right)+\left(12z-36\right)$$

Factor out z in the first and 12 in the second group.

$$z\left(z-3\right)+12\left(z-3\right)$$

Factor out common term z-3 by using distributive property.

$$\left(z-3\right)\left(z+12\right)$$

Evaluate

$$\left(z-3\right)\left(z+12\right)$$

Solution Steps

Combine 12z and -3z to get 9z.

$$z^{2}+9z-36$$