Solve 8z-6-2z^2=

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

Factor

$$2\left(1-z\right)\left(z-3\right)$$

Solution Steps

Factor out 2.

$$2\left(4z-3-z^{2}\right)$$

Consider 4z-3-z^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-z^{2}+4z-3$$

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

$$a+b=4 ab=-\left(-3\right)=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=3 b=1$$

Rewrite -z^{2}+4z-3 as \left(-z^{2}+3z\right)+\left(z-3\right).

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

Factor out -z in -z^{2}+3z.

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

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

$$\left(z-3\right)\left(-z+1\right)$$

Rewrite the complete factored expression.

$$2\left(z-3\right)\left(-z+1\right)$$

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.

$$-2z^{2}+8z-6=0$$

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.

$$z=\frac{-8±\sqrt{8^{2}-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}$$

Square 8.

$$z=\frac{-8±\sqrt{64-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$z=\frac{-8±\sqrt{64+8\left(-6\right)}}{2\left(-2\right)}$$

Multiply 8 times -6.

$$z=\frac{-8±\sqrt{64-48}}{2\left(-2\right)}$$

Add 64 to -48.

$$z=\frac{-8±\sqrt{16}}{2\left(-2\right)}$$

Take the square root of 16.

$$z=\frac{-8±4}{2\left(-2\right)}$$

Multiply 2 times -2.

$$z=\frac{-8±4}{-4}$$

Now solve the equation z=\frac{-8±4}{-4} when ± is plus. Add -8 to 4.

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

Divide -4 by -4.

$$z=1$$

Now solve the equation z=\frac{-8±4}{-4} when ± is minus. Subtract 4 from -8.

$$z=-\frac{12}{-4}$$

Divide -12 by -4.

$$z=3$$

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

$$-2z^{2}+8z-6=-2\left(z-1\right)\left(z-3\right)$$

Evaluate

$$2\left(1-z\right)\left(z-3\right)$$