Solve z^2-12z+32

$$z ^ { 2 } - 12 z + 32$$

Factor

$$\left(z-8\right)\left(z-4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz+32. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\times 32=32$$

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 32.

$$-1,-32 -2,-16 -4,-8$$

Calculate the sum for each pair.

$$-1-32=-33 -2-16=-18 -4-8=-12$$

The solution is the pair that gives sum -12.

$$a=-8 b=-4$$

Rewrite z^{2}-12z+32 as \left(z^{2}-8z\right)+\left(-4z+32\right).

$$\left(z^{2}-8z\right)+\left(-4z+32\right)$$

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

$$z\left(z-8\right)-4\left(z-8\right)$$

Factor out common term z-8 by using distributive property.

$$\left(z-8\right)\left(z-4\right)$$

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.

$$z^{2}-12z+32=0$$

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.

$$z=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 32}}{2}$$

Square -12.

$$z=\frac{-\left(-12\right)±\sqrt{144-4\times 32}}{2}$$

Multiply -4 times 32.

$$z=\frac{-\left(-12\right)±\sqrt{144-128}}{2}$$

Add 144 to -128.

$$z=\frac{-\left(-12\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$z=\frac{-\left(-12\right)±4}{2}$$

The opposite of -12 is 12.

$$z=\frac{12±4}{2}$$

Now solve the equation z=\frac{12±4}{2} when ± is plus. Add 12 to 4.

$$z=\frac{16}{2}$$

Divide 16 by 2.

$$z=8$$

Now solve the equation z=\frac{12±4}{2} when ± is minus. Subtract 4 from 12.

$$z=\frac{8}{2}$$

Divide 8 by 2.

$$z=4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and 4 for x_{2}.

$$z^{2}-12z+32=\left(z-8\right)\left(z-4\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -12x +32 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 12 rs = 32$$

Two numbers r and s sum up to 12 exactly when the average of the two numbers is \frac{1}{2}*12 = 6. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 6 - u s = 6 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 32

$$(6 - u) (6 + u) = 32$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$36 - u^2 = 32$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = 32-36 = -4$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =6 - 2 = 4 s = 6 + 2 = 8$$

Evaluate

$$\left(z-8\right)\left(z-4\right)$$