Solve z^2-12z-45

$$z ^ { 2 } - 12 z - 45$$

Factor

$$\left(z-15\right)\left(z+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz-45. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\left(-45\right)=-45$$

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 -45.

$$1,-45 3,-15 5,-9$$

Calculate the sum for each pair.

$$1-45=-44 3-15=-12 5-9=-4$$

The solution is the pair that gives sum -12.

$$a=-15 b=3$$

Rewrite z^{2}-12z-45 as \left(z^{2}-15z\right)+\left(3z-45\right).

$$\left(z^{2}-15z\right)+\left(3z-45\right)$$

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

$$z\left(z-15\right)+3\left(z-15\right)$$

Factor out common term z-15 by using distributive property.

$$\left(z-15\right)\left(z+3\right)$$

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.

$$z^{2}-12z-45=0$$

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.

$$z=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-45\right)}}{2}$$

Square -12.

$$z=\frac{-\left(-12\right)±\sqrt{144-4\left(-45\right)}}{2}$$

Multiply -4 times -45.

$$z=\frac{-\left(-12\right)±\sqrt{144+180}}{2}$$

Add 144 to 180.

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

Take the square root of 324.

$$z=\frac{-\left(-12\right)±18}{2}$$

The opposite of -12 is 12.

$$z=\frac{12±18}{2}$$

Now solve the equation z=\frac{12±18}{2} when ± is plus. Add 12 to 18.

$$z=\frac{30}{2}$$

Divide 30 by 2.

$$z=15$$

Now solve the equation z=\frac{12±18}{2} when ± is minus. Subtract 18 from 12.

$$z=-\frac{6}{2}$$

Divide -6 by 2.

$$z=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 15 for x_{1} and -3 for x_{2}.

$$z^{2}-12z-45=\left(z-15\right)\left(z-\left(-3\right)\right)$$

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

$$z^{2}-12z-45=\left(z-15\right)\left(z+3\right)$$

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.

$$x ^ 2 -12x -45 = 0$$

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

$$r + s = 12 rs = -45$$

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 = -45

$$(6 - u) (6 + u) = -45$$

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

$$36 - u^2 = -45$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = -45-36 = -81$$

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

$$u^2 = 81 u = \pm\sqrt{81} = \pm 9$$

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

$$r =6 - 9 = -3 s = 6 + 9 = 15$$

Evaluate

$$\left(z-15\right)\left(z+3\right)$$