Solve 18z-24-3z^2

$$18 z - 24 - 3 z ^ { 2 }$$

Factor

$$3\left(2-z\right)\left(z-4\right)$$

Solution Steps

Factor out 3.

$$3\left(6z-8-z^{2}\right)$$

Consider 6z-8-z^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-z^{2}+6z-8$$

Factor the expression by grouping. First, the expression needs to be rewritten as -z^{2}+az+bz-8. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-8\right)=8$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=4 b=2$$

Rewrite -z^{2}+6z-8 as \left(-z^{2}+4z\right)+\left(2z-8\right).

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

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

$$-z\left(z-4\right)+2\left(z-4\right)$$

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

$$\left(z-4\right)\left(-z+2\right)$$

Rewrite the complete factored expression.

$$3\left(z-4\right)\left(-z+2\right)$$

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.

$$-3z^{2}+18z-24=0$$

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.

$$z=\frac{-18±\sqrt{18^{2}-4\left(-3\right)\left(-24\right)}}{2\left(-3\right)}$$

Square 18.

$$z=\frac{-18±\sqrt{324-4\left(-3\right)\left(-24\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$z=\frac{-18±\sqrt{324+12\left(-24\right)}}{2\left(-3\right)}$$

Multiply 12 times -24.

$$z=\frac{-18±\sqrt{324-288}}{2\left(-3\right)}$$

Add 324 to -288.

$$z=\frac{-18±\sqrt{36}}{2\left(-3\right)}$$

Take the square root of 36.

$$z=\frac{-18±6}{2\left(-3\right)}$$

Multiply 2 times -3.

$$z=\frac{-18±6}{-6}$$

Now solve the equation z=\frac{-18±6}{-6} when ± is plus. Add -18 to 6.

$$z=-\frac{12}{-6}$$

Divide -12 by -6.

$$z=2$$

Now solve the equation z=\frac{-18±6}{-6} when ± is minus. Subtract 6 from -18.

$$z=-\frac{24}{-6}$$

Divide -24 by -6.

$$z=4$$

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

$$-3z^{2}+18z-24=-3\left(z-2\right)\left(z-4\right)$$

Evaluate

$$3\left(2-z\right)\left(z-4\right)$$