Solve p^2-15p+36

$$p ^ { 2 } - 15 p + 36$$

Factor

$$\left(p-12\right)\left(p-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp+36. To find a and b, set up a system to be solved.

$$a+b=-15 ab=1\times 36=36$$

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 36.

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

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -15.

$$a=-12 b=-3$$

Rewrite p^{2}-15p+36 as \left(p^{2}-12p\right)+\left(-3p+36\right).

$$\left(p^{2}-12p\right)+\left(-3p+36\right)$$

Factor out p in the first and -3 in the second group.

$$p\left(p-12\right)-3\left(p-12\right)$$

Factor out common term p-12 by using distributive property.

$$\left(p-12\right)\left(p-3\right)$$

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.

$$p^{2}-15p+36=0$$

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.

$$p=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 36}}{2}$$

Square -15.

$$p=\frac{-\left(-15\right)±\sqrt{225-4\times 36}}{2}$$

Multiply -4 times 36.

$$p=\frac{-\left(-15\right)±\sqrt{225-144}}{2}$$

Add 225 to -144.

$$p=\frac{-\left(-15\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$p=\frac{-\left(-15\right)±9}{2}$$

The opposite of -15 is 15.

$$p=\frac{15±9}{2}$$

Now solve the equation p=\frac{15±9}{2} when ± is plus. Add 15 to 9.

$$p=\frac{24}{2}$$

Divide 24 by 2.

$$p=12$$

Now solve the equation p=\frac{15±9}{2} when ± is minus. Subtract 9 from 15.

$$p=\frac{6}{2}$$

Divide 6 by 2.

$$p=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 12 for x_{1} and 3 for x_{2}.

$$p^{2}-15p+36=\left(p-12\right)\left(p-3\right)$$

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.

$$x ^ 2 -15x +36 = 0$$

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

$$r + s = 15 rs = 36$$

Two numbers r and s sum up to 15 exactly when the average of the two numbers is \frac{1}{2}*15 = \frac{15}{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>

$$r = \frac{15}{2} - u s = \frac{15}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 36

$$(\frac{15}{2} - u) (\frac{15}{2} + u) = 36$$

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

$$\frac{225}{4} - u^2 = 36$$

Simplify the expression by subtracting \frac{225}{4} on both sides

$$-u^2 = 36-\frac{225}{4} = -\frac{81}{4}$$

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

$$u^2 = \frac{81}{4} u = \pm\sqrt{\frac{81}{4}} = \pm \frac{9}{2}$$

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

$$r =\frac{15}{2} - \frac{9}{2} = 3 s = \frac{15}{2} + \frac{9}{2} = 12$$

Evaluate

$$\left(p-12\right)\left(p-3\right)$$