Solve x^2+30x-99

$${ x }^{ 2 } +30x-99$$

Factor

$$\left(x-3\right)\left(x+33\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-99. To find a and b, set up a system to be solved.

$$a+b=30 ab=1\left(-99\right)=-99$$

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 -99.

$$-1,99 -3,33 -9,11$$

Calculate the sum for each pair.

$$-1+99=98 -3+33=30 -9+11=2$$

The solution is the pair that gives sum 30.

$$a=-3 b=33$$

Rewrite x^{2}+30x-99 as \left(x^{2}-3x\right)+\left(33x-99\right).

$$\left(x^{2}-3x\right)+\left(33x-99\right)$$

Factor out x in the first and 33 in the second group.

$$x\left(x-3\right)+33\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+33\right)$$

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.

$$x^{2}+30x-99=0$$

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.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-99\right)}}{2}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\left(-99\right)}}{2}$$

Multiply -4 times -99.

$$x=\frac{-30±\sqrt{900+396}}{2}$$

Add 900 to 396.

$$x=\frac{-30±\sqrt{1296}}{2}$$

Take the square root of 1296.

$$x=\frac{-30±36}{2}$$

Now solve the equation x=\frac{-30±36}{2} when ± is plus. Add -30 to 36.

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

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-30±36}{2} when ± is minus. Subtract 36 from -30.

$$x=-\frac{66}{2}$$

Divide -66 by 2.

$$x=-33$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and -33 for x_{2}.

$$x^{2}+30x-99=\left(x-3\right)\left(x-\left(-33\right)\right)$$

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

$$x^{2}+30x-99=\left(x-3\right)\left(x+33\right)$$

Evaluate

$$\left(x-3\right)\left(x+33\right)$$