Solve a^2-16a-132

$$a ^ { 2 } - 16 a - 132$$

Factor

$$\left(a-22\right)\left(a+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-132. To find p and q, set up a system to be solved.

$$p+q=-16 pq=1\left(-132\right)=-132$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -132.

$$1,-132 2,-66 3,-44 4,-33 6,-22 11,-12$$

Calculate the sum for each pair.

$$1-132=-131 2-66=-64 3-44=-41 4-33=-29 6-22=-16 11-12=-1$$

The solution is the pair that gives sum -16.

$$p=-22 q=6$$

Rewrite a^{2}-16a-132 as \left(a^{2}-22a\right)+\left(6a-132\right).

$$\left(a^{2}-22a\right)+\left(6a-132\right)$$

Factor out a in the first and 6 in the second group.

$$a\left(a-22\right)+6\left(a-22\right)$$

Factor out common term a-22 by using distributive property.

$$\left(a-22\right)\left(a+6\right)$$

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.

$$a^{2}-16a-132=0$$

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.

$$a=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\left(-132\right)}}{2}$$

Square -16.

$$a=\frac{-\left(-16\right)±\sqrt{256-4\left(-132\right)}}{2}$$

Multiply -4 times -132.

$$a=\frac{-\left(-16\right)±\sqrt{256+528}}{2}$$

Add 256 to 528.

$$a=\frac{-\left(-16\right)±\sqrt{784}}{2}$$

Take the square root of 784.

$$a=\frac{-\left(-16\right)±28}{2}$$

The opposite of -16 is 16.

$$a=\frac{16±28}{2}$$

Now solve the equation a=\frac{16±28}{2} when ± is plus. Add 16 to 28.

$$a=\frac{44}{2}$$

Divide 44 by 2.

$$a=22$$

Now solve the equation a=\frac{16±28}{2} when ± is minus. Subtract 28 from 16.

$$a=-\frac{12}{2}$$

Divide -12 by 2.

$$a=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 22 for x_{1} and -6 for x_{2}.

$$a^{2}-16a-132=\left(a-22\right)\left(a-\left(-6\right)\right)$$

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

$$a^{2}-16a-132=\left(a-22\right)\left(a+6\right)$$

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.

$$x ^ 2 -16x -132 = 0$$

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

$$r + s = 16 rs = -132$$

Two numbers r and s sum up to 16 exactly when the average of the two numbers is \frac{1}{2}*16 = 8. 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>

$$r = 8 - u s = 8 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -132

$$(8 - u) (8 + u) = -132$$

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

$$64 - u^2 = -132$$

Simplify the expression by subtracting 64 on both sides

$$-u^2 = -132-64 = -196$$

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

$$u^2 = 196 u = \pm\sqrt{196} = \pm 14$$

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

$$r =8 - 14 = -6 s = 8 + 14 = 22$$

Evaluate

$$\left(a-22\right)\left(a+6\right)$$