Solve x^2+x-132=*

$${ x }^{ 2 } +x-132= \times$$

Factor

$$\left(x-11\right)\left(x+12\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-132. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-132\right)=-132$$

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 -132.

$$-1,132 -2,66 -3,44 -4,33 -6,22 -11,12$$

Calculate the sum for each pair.

$$-1+132=131 -2+66=64 -3+44=41 -4+33=29 -6+22=16 -11+12=1$$

The solution is the pair that gives sum 1.

$$a=-11 b=12$$

Rewrite x^{2}+x-132 as \left(x^{2}-11x\right)+\left(12x-132\right).

$$\left(x^{2}-11x\right)+\left(12x-132\right)$$

Factor out x in the first and 12 in the second group.

$$x\left(x-11\right)+12\left(x-11\right)$$

Factor out common term x-11 by using distributive property.

$$\left(x-11\right)\left(x+12\right)$$

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.

$$x^{2}+x-132=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-132\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-132\right)}}{2}$$

Multiply -4 times -132.

$$x=\frac{-1±\sqrt{1+528}}{2}$$

Add 1 to 528.

$$x=\frac{-1±\sqrt{529}}{2}$$

Take the square root of 529.

$$x=\frac{-1±23}{2}$$

Now solve the equation x=\frac{-1±23}{2} when ± is plus. Add -1 to 23.

$$x=\frac{22}{2}$$

Divide 22 by 2.

$$x=11$$

Now solve the equation x=\frac{-1±23}{2} when ± is minus. Subtract 23 from -1.

$$x=-\frac{24}{2}$$

Divide -24 by 2.

$$x=-12$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 11 for x_{1} and -12 for x_{2}.

$$x^{2}+x-132=\left(x-11\right)\left(x-\left(-12\right)\right)$$

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

$$x^{2}+x-132=\left(x-11\right)\left(x+12\right)$$

Evaluate

$$\left(x-11\right)\left(x+12\right)$$