Solve 3x-18+x^2

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

Factor

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

Solution Steps

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

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

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 positive, the positive number has greater absolute value than the negative. 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.

$$a=-3 b=6$$

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

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

Factor out x in the first and 6 in the second group.

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

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

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

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.

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

Square 3.

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

Multiply -4 times -18.

$$x=\frac{-3±\sqrt{9+72}}{2}$$

Add 9 to 72.

$$x=\frac{-3±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-3±9}{2}$$

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

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

Divide 6 by 2.

$$x=3$$

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

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

Divide -12 by 2.

$$x=-6$$

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 -6 for x_{2}.

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

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

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

Evaluate

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