Solve -22z+6z^2+12

$$- 22 z + 6 z ^ { 2 } + 12$$

Factor

$$2\left(z-3\right)\left(3z-2\right)$$

Solution Steps

Factor out 2.

$$2\left(-11z+3z^{2}+6\right)$$

Consider -11z+3z^{2}+6. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$3z^{2}-11z+6$$

Factor the expression by grouping. First, the expression needs to be rewritten as 3z^{2}+az+bz+6. To find a and b, set up a system to be solved.

$$a+b=-11 ab=3\times 6=18$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -11.

$$a=-9 b=-2$$

Rewrite 3z^{2}-11z+6 as \left(3z^{2}-9z\right)+\left(-2z+6\right).

$$\left(3z^{2}-9z\right)+\left(-2z+6\right)$$

Factor out 3z in the first and -2 in the second group.

$$3z\left(z-3\right)-2\left(z-3\right)$$

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

$$\left(z-3\right)\left(3z-2\right)$$

Rewrite the complete factored expression.

$$2\left(z-3\right)\left(3z-2\right)$$

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.

$$6z^{2}-22z+12=0$$

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.

$$z=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 6\times 12}}{2\times 6}$$

Square -22.

$$z=\frac{-\left(-22\right)±\sqrt{484-4\times 6\times 12}}{2\times 6}$$

Multiply -4 times 6.

$$z=\frac{-\left(-22\right)±\sqrt{484-24\times 12}}{2\times 6}$$

Multiply -24 times 12.

$$z=\frac{-\left(-22\right)±\sqrt{484-288}}{2\times 6}$$

Add 484 to -288.

$$z=\frac{-\left(-22\right)±\sqrt{196}}{2\times 6}$$

Take the square root of 196.

$$z=\frac{-\left(-22\right)±14}{2\times 6}$$

The opposite of -22 is 22.

$$z=\frac{22±14}{2\times 6}$$

Multiply 2 times 6.

$$z=\frac{22±14}{12}$$

Now solve the equation z=\frac{22±14}{12} when ± is plus. Add 22 to 14.

$$z=\frac{36}{12}$$

Divide 36 by 12.

$$z=3$$

Now solve the equation z=\frac{22±14}{12} when ± is minus. Subtract 14 from 22.

$$z=\frac{8}{12}$$

Reduce the fraction \frac{8}{12} to lowest terms by extracting and canceling out 4.

$$z=\frac{2}{3}$$

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

$$6z^{2}-22z+12=6\left(z-3\right)\left(z-\frac{2}{3}\right)$$

Subtract \frac{2}{3} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6z^{2}-22z+12=6\left(z-3\right)\times \frac{3z-2}{3}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6z^{2}-22z+12=2\left(z-3\right)\left(3z-2\right)$$

Evaluate

$$2\left(z-3\right)\left(3z-2\right)$$