Solve 12z^2-7z-12

$$12 { z }^{ 2 } -7z-12$$

Factor

$$\left(3z-4\right)\left(4z+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12z^{2}+az+bz-12. To find a and b, set up a system to be solved.

$$a+b=-7 ab=12\left(-12\right)=-144$$

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 -144.

$$1,-144 2,-72 3,-48 4,-36 6,-24 8,-18 9,-16 12,-12$$

Calculate the sum for each pair.

$$1-144=-143 2-72=-70 3-48=-45 4-36=-32 6-24=-18 8-18=-10 9-16=-7 12-12=0$$

The solution is the pair that gives sum -7.

$$a=-16 b=9$$

Rewrite 12z^{2}-7z-12 as \left(12z^{2}-16z\right)+\left(9z-12\right).

$$\left(12z^{2}-16z\right)+\left(9z-12\right)$$

Factor out 4z in the first and 3 in the second group.

$$4z\left(3z-4\right)+3\left(3z-4\right)$$

Factor out common term 3z-4 by using distributive property.

$$\left(3z-4\right)\left(4z+3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$12z^{2}-7z-12=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$z=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12\left(-12\right)}}{2\times 12}$$

Square -7.

$$z=\frac{-\left(-7\right)±\sqrt{49-4\times 12\left(-12\right)}}{2\times 12}$$

Multiply -4 times 12.

$$z=\frac{-\left(-7\right)±\sqrt{49-48\left(-12\right)}}{2\times 12}$$

Multiply -48 times -12.

$$z=\frac{-\left(-7\right)±\sqrt{49+576}}{2\times 12}$$

Add 49 to 576.

$$z=\frac{-\left(-7\right)±\sqrt{625}}{2\times 12}$$

Take the square root of 625.

$$z=\frac{-\left(-7\right)±25}{2\times 12}$$

The opposite of -7 is 7.

$$z=\frac{7±25}{2\times 12}$$

Multiply 2 times 12.

$$z=\frac{7±25}{24}$$

Now solve the equation z=\frac{7±25}{24} when ± is plus. Add 7 to 25.

$$z=\frac{32}{24}$$

Reduce the fraction \frac{32}{24} to lowest terms by extracting and canceling out 8.

$$z=\frac{4}{3}$$

Now solve the equation z=\frac{7±25}{24} when ± is minus. Subtract 25 from 7.

$$z=-\frac{18}{24}$$

Reduce the fraction \frac{-18}{24} to lowest terms by extracting and canceling out 6.

$$z=-\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and -\frac{3}{4} for x_{2}.

$$12z^{2}-7z-12=12\left(z-\frac{4}{3}\right)\left(z-\left(-\frac{3}{4}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$12z^{2}-7z-12=12\left(z-\frac{4}{3}\right)\left(z+\frac{3}{4}\right)$$

Subtract \frac{4}{3} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12z^{2}-7z-12=12\times \frac{3z-4}{3}\left(z+\frac{3}{4}\right)$$

Add \frac{3}{4} to z by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12z^{2}-7z-12=12\times \frac{3z-4}{3}\times \frac{4z+3}{4}$$

Multiply \frac{3z-4}{3} times \frac{4z+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12z^{2}-7z-12=12\times \frac{\left(3z-4\right)\left(4z+3\right)}{3\times 4}$$

Multiply 3 times 4.

$$12z^{2}-7z-12=12\times \frac{\left(3z-4\right)\left(4z+3\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12z^{2}-7z-12=\left(3z-4\right)\left(4z+3\right)$$

Evaluate

$$\left(3z-4\right)\left(4z+3\right)$$