Solve p^2-31p+108

$$p ^ { 2 } - 31 p + 108$$

Factor

$$\left(p-27\right)\left(p-4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp+108. To find a and b, set up a system to be solved.

$$a+b=-31 ab=1\times 108=108$$

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 108.

$$-1,-108 -2,-54 -3,-36 -4,-27 -6,-18 -9,-12$$

Calculate the sum for each pair.

$$-1-108=-109 -2-54=-56 -3-36=-39 -4-27=-31 -6-18=-24 -9-12=-21$$

The solution is the pair that gives sum -31.

$$a=-27 b=-4$$

Rewrite p^{2}-31p+108 as \left(p^{2}-27p\right)+\left(-4p+108\right).

$$\left(p^{2}-27p\right)+\left(-4p+108\right)$$

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

$$p\left(p-27\right)-4\left(p-27\right)$$

Factor out common term p-27 by using distributive property.

$$\left(p-27\right)\left(p-4\right)$$

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.

$$p^{2}-31p+108=0$$

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.

$$p=\frac{-\left(-31\right)±\sqrt{\left(-31\right)^{2}-4\times 108}}{2}$$

Square -31.

$$p=\frac{-\left(-31\right)±\sqrt{961-4\times 108}}{2}$$

Multiply -4 times 108.

$$p=\frac{-\left(-31\right)±\sqrt{961-432}}{2}$$

Add 961 to -432.

$$p=\frac{-\left(-31\right)±\sqrt{529}}{2}$$

Take the square root of 529.

$$p=\frac{-\left(-31\right)±23}{2}$$

The opposite of -31 is 31.

$$p=\frac{31±23}{2}$$

Now solve the equation p=\frac{31±23}{2} when ± is plus. Add 31 to 23.

$$p=\frac{54}{2}$$

Divide 54 by 2.

$$p=27$$

Now solve the equation p=\frac{31±23}{2} when ± is minus. Subtract 23 from 31.

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

Divide 8 by 2.

$$p=4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 27 for x_{1} and 4 for x_{2}.

$$p^{2}-31p+108=\left(p-27\right)\left(p-4\right)$$

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.

$$x ^ 2 -31x +108 = 0$$

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

$$r + s = 31 rs = 108$$

Two numbers r and s sum up to 31 exactly when the average of the two numbers is \frac{1}{2}*31 = \frac{31}{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>

$$r = \frac{31}{2} - u s = \frac{31}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 108

$$(\frac{31}{2} - u) (\frac{31}{2} + u) = 108$$

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

$$\frac{961}{4} - u^2 = 108$$

Simplify the expression by subtracting \frac{961}{4} on both sides

$$-u^2 = 108-\frac{961}{4} = -\frac{529}{4}$$

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

$$u^2 = \frac{529}{4} u = \pm\sqrt{\frac{529}{4}} = \pm \frac{23}{2}$$

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

$$r =\frac{31}{2} - \frac{23}{2} = 4 s = \frac{31}{2} + \frac{23}{2} = 27$$

Evaluate

$$\left(p-27\right)\left(p-4\right)$$