Solve x^2+12x-45

$${ x }^{ 2 } +12x-45$$

Factor

$$\left(x-3\right)\left(x+15\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-45. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\left(-45\right)=-45$$

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 -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 12.

$$a=-3 b=15$$

Rewrite x^{2}+12x-45 as \left(x^{2}-3x\right)+\left(15x-45\right).

$$\left(x^{2}-3x\right)+\left(15x-45\right)$$

Factor out x in the first and 15 in the second group.

$$x\left(x-3\right)+15\left(x-3\right)$$

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

$$\left(x-3\right)\left(x+15\right)$$

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.

$$x^{2}+12x-45=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-45\right)}}{2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-45\right)}}{2}$$

Multiply -4 times -45.

$$x=\frac{-12±\sqrt{144+180}}{2}$$

Add 144 to 180.

$$x=\frac{-12±\sqrt{324}}{2}$$

Take the square root of 324.

$$x=\frac{-12±18}{2}$$

Now solve the equation x=\frac{-12±18}{2} when ± is plus. Add -12 to 18.

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

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-12±18}{2} when ± is minus. Subtract 18 from -12.

$$x=-\frac{30}{2}$$

Divide -30 by 2.

$$x=-15$$

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 -15 for x_{2}.

$$x^{2}+12x-45=\left(x-3\right)\left(x-\left(-15\right)\right)$$

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

$$x^{2}+12x-45=\left(x-3\right)\left(x+15\right)$$

Evaluate

$$\left(x-3\right)\left(x+15\right)$$