Solve {r}{x^2-3x-18}{-0}

$$\left. \begin{array} { r } { x ^ { 2 } - 3 x - 18 } \\ { - 0 } \end{array} \right.$$

Evaluate

$$\left(x-6\right)\left(x+3\right)$$

Solution Steps

Multiply -1 and 0 to get 0.

$$x^{2}-3x-18+0$$

Add -18 and 0 to get -18.

$$x^{2}-3x-18$$

Factor

$$\left(x-6\right)\left(x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-18\right)=-18$$

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 -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -3.

$$a=-6 b=3$$

Rewrite x^{2}-3x-18 as \left(x^{2}-6x\right)+\left(3x-18\right).

$$\left(x^{2}-6x\right)+\left(3x-18\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-6\right)+3\left(x-6\right)$$

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

$$\left(x-6\right)\left(x+3\right)$$

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.

$$x^{2}-3x-18=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-18\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$x=\frac{-\left(-3\right)±\sqrt{9+72}}{2}$$

Add 9 to 72.

$$x=\frac{-\left(-3\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-\left(-3\right)±9}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±9}{2}$$

Now solve the equation x=\frac{3±9}{2} when ± is plus. Add 3 to 9.

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

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{3±9}{2} when ± is minus. Subtract 9 from 3.

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

Divide -6 by 2.

$$x=-3$$

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 -3 for x_{2}.

$$x^{2}-3x-18=\left(x-6\right)\left(x-\left(-3\right)\right)$$

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

$$x^{2}-3x-18=\left(x-6\right)\left(x+3\right)$$