Solve n^2-28-12n

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

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

Factor the expression by grouping. First, the expression needs to be rewritten as n^{2}+an+bn-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 n^{2}-12n-28 as \left(n^{2}-14n\right)+\left(2n-28\right).

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

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

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

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

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

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.

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

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.

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

Square -12.

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

Multiply -4 times -28.

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

Add 144 to 112.

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

Take the square root of 256.

$$n=\frac{-\left(-12\right)±16}{2}$$

The opposite of -12 is 12.

$$n=\frac{12±16}{2}$$

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

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

Divide 28 by 2.

$$n=14$$

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

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

Divide -4 by 2.

$$n=-2$$

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}.

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

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

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

Evaluate

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