Solve p^2-4p-117

$$p ^ { 2 } - 4 p - 117$$

Factor

$$\left(p-13\right)\left(p+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp-117. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-117\right)=-117$$

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 -117.

$$1,-117 3,-39 9,-13$$

Calculate the sum for each pair.

$$1-117=-116 3-39=-36 9-13=-4$$

The solution is the pair that gives sum -4.

$$a=-13 b=9$$

Rewrite p^{2}-4p-117 as \left(p^{2}-13p\right)+\left(9p-117\right).

$$\left(p^{2}-13p\right)+\left(9p-117\right)$$

Factor out p in the first and 9 in the second group.

$$p\left(p-13\right)+9\left(p-13\right)$$

Factor out common term p-13 by using distributive property.

$$\left(p-13\right)\left(p+9\right)$$

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.

$$p^{2}-4p-117=0$$

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.

$$p=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-117\right)}}{2}$$

Square -4.

$$p=\frac{-\left(-4\right)±\sqrt{16-4\left(-117\right)}}{2}$$

Multiply -4 times -117.

$$p=\frac{-\left(-4\right)±\sqrt{16+468}}{2}$$

Add 16 to 468.

$$p=\frac{-\left(-4\right)±\sqrt{484}}{2}$$

Take the square root of 484.

$$p=\frac{-\left(-4\right)±22}{2}$$

The opposite of -4 is 4.

$$p=\frac{4±22}{2}$$

Now solve the equation p=\frac{4±22}{2} when ± is plus. Add 4 to 22.

$$p=\frac{26}{2}$$

Divide 26 by 2.

$$p=13$$

Now solve the equation p=\frac{4±22}{2} when ± is minus. Subtract 22 from 4.

$$p=-\frac{18}{2}$$

Divide -18 by 2.

$$p=-9$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 13 for x_{1} and -9 for x_{2}.

$$p^{2}-4p-117=\left(p-13\right)\left(p-\left(-9\right)\right)$$

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

$$p^{2}-4p-117=\left(p-13\right)\left(p+9\right)$$

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.

$$x ^ 2 -4x -117 = 0$$

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

$$r + s = 4 rs = -117$$

Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 2. 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>

$$r = 2 - u s = 2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -117

$$(2 - u) (2 + u) = -117$$

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

$$4 - u^2 = -117$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -117-4 = -121$$

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

$$u^2 = 121 u = \pm\sqrt{121} = \pm 11$$

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

$$r =2 - 11 = -9 s = 2 + 11 = 13$$

Evaluate

$$\left(p-13\right)\left(p+9\right)$$