Solve p^2-12p-28

$$p ^ { 2 } - 12 p - 28$$

Factor

$$\left(p-14\right)\left(p+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp-28. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\left(-28\right)=-28$$

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 -28.

$$1,-28 2,-14 4,-7$$

Calculate the sum for each pair.

$$1-28=-27 2-14=-12 4-7=-3$$

The solution is the pair that gives sum -12.

$$a=-14 b=2$$

Rewrite p^{2}-12p-28 as \left(p^{2}-14p\right)+\left(2p-28\right).

$$\left(p^{2}-14p\right)+\left(2p-28\right)$$

Factor out p in the first and 2 in the second group.

$$p\left(p-14\right)+2\left(p-14\right)$$

Factor out common term p-14 by using distributive property.

$$\left(p-14\right)\left(p+2\right)$$

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.

$$p^{2}-12p-28=0$$

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.

$$p=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-28\right)}}{2}$$

Square -12.

$$p=\frac{-\left(-12\right)±\sqrt{144-4\left(-28\right)}}{2}$$

Multiply -4 times -28.

$$p=\frac{-\left(-12\right)±\sqrt{144+112}}{2}$$

Add 144 to 112.

$$p=\frac{-\left(-12\right)±\sqrt{256}}{2}$$

Take the square root of 256.

$$p=\frac{-\left(-12\right)±16}{2}$$

The opposite of -12 is 12.

$$p=\frac{12±16}{2}$$

Now solve the equation p=\frac{12±16}{2} when ± is plus. Add 12 to 16.

$$p=\frac{28}{2}$$

Divide 28 by 2.

$$p=14$$

Now solve the equation p=\frac{12±16}{2} when ± is minus. Subtract 16 from 12.

$$p=-\frac{4}{2}$$

Divide -4 by 2.

$$p=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 14 for x_{1} and -2 for x_{2}.

$$p^{2}-12p-28=\left(p-14\right)\left(p-\left(-2\right)\right)$$

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

$$p^{2}-12p-28=\left(p-14\right)\left(p+2\right)$$

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.

$$x ^ 2 -12x -28 = 0$$

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

$$r + s = 12 rs = -28$$

Two numbers r and s sum up to 12 exactly when the average of the two numbers is \frac{1}{2}*12 = 6. 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>

$$r = 6 - u s = 6 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -28

$$(6 - u) (6 + u) = -28$$

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

$$36 - u^2 = -28$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = -28-36 = -64$$

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

$$u^2 = 64 u = \pm\sqrt{64} = \pm 8$$

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

$$r =6 - 8 = -2 s = 6 + 8 = 14$$

Evaluate

$$\left(p-14\right)\left(p+2\right)$$