Solve p^2+18p-144

$$p ^ { 2 } + 18 p - 144$$

Factor

$$\left(p-6\right)\left(p+24\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp-144. To find a and b, set up a system to be solved.

$$a+b=18 ab=1\left(-144\right)=-144$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -144.

$$-1,144 -2,72 -3,48 -4,36 -6,24 -8,18 -9,16 -12,12$$

Calculate the sum for each pair.

$$-1+144=143 -2+72=70 -3+48=45 -4+36=32 -6+24=18 -8+18=10 -9+16=7 -12+12=0$$

The solution is the pair that gives sum 18.

$$a=-6 b=24$$

Rewrite p^{2}+18p-144 as \left(p^{2}-6p\right)+\left(24p-144\right).

$$\left(p^{2}-6p\right)+\left(24p-144\right)$$

Factor out p in the first and 24 in the second group.

$$p\left(p-6\right)+24\left(p-6\right)$$

Factor out common term p-6 by using distributive property.

$$\left(p-6\right)\left(p+24\right)$$

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.

$$p^{2}+18p-144=0$$

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.

$$p=\frac{-18±\sqrt{18^{2}-4\left(-144\right)}}{2}$$

Square 18.

$$p=\frac{-18±\sqrt{324-4\left(-144\right)}}{2}$$

Multiply -4 times -144.

$$p=\frac{-18±\sqrt{324+576}}{2}$$

Add 324 to 576.

$$p=\frac{-18±\sqrt{900}}{2}$$

Take the square root of 900.

$$p=\frac{-18±30}{2}$$

Now solve the equation p=\frac{-18±30}{2} when ± is plus. Add -18 to 30.

$$p=\frac{12}{2}$$

Divide 12 by 2.

$$p=6$$

Now solve the equation p=\frac{-18±30}{2} when ± is minus. Subtract 30 from -18.

$$p=-\frac{48}{2}$$

Divide -48 by 2.

$$p=-24$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and -24 for x_{2}.

$$p^{2}+18p-144=\left(p-6\right)\left(p-\left(-24\right)\right)$$

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

$$p^{2}+18p-144=\left(p-6\right)\left(p+24\right)$$

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.

$$x ^ 2 +18x -144 = 0$$

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

$$r + s = -18 rs = -144$$

Two numbers r and s sum up to -18 exactly when the average of the two numbers is \frac{1}{2}*-18 = -9. 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>

$$r = -9 - u s = -9 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -144

$$(-9 - u) (-9 + u) = -144$$

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

$$81 - u^2 = -144$$

Simplify the expression by subtracting 81 on both sides

$$-u^2 = -144-81 = -225$$

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

$$u^2 = 225 u = \pm\sqrt{225} = \pm 15$$

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

$$r =-9 - 15 = -24 s = -9 + 15 = 6$$

Evaluate

$$\left(p-6\right)\left(p+24\right)$$