Solve x^2-6x-55

$${ x }^{ 2 } -6x-55$$

Factor

$$\left(x-11\right)\left(x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-55. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-55\right)=-55$$

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 -55.

$$1,-55 5,-11$$

Calculate the sum for each pair.

$$1-55=-54 5-11=-6$$

The solution is the pair that gives sum -6.

$$a=-11 b=5$$

Rewrite x^{2}-6x-55 as \left(x^{2}-11x\right)+\left(5x-55\right).

$$\left(x^{2}-11x\right)+\left(5x-55\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-11\right)+5\left(x-11\right)$$

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

$$\left(x-11\right)\left(x+5\right)$$

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.

$$x^{2}-6x-55=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-55\right)}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-55\right)}}{2}$$

Multiply -4 times -55.

$$x=\frac{-\left(-6\right)±\sqrt{36+220}}{2}$$

Add 36 to 220.

$$x=\frac{-\left(-6\right)±\sqrt{256}}{2}$$

Take the square root of 256.

$$x=\frac{-\left(-6\right)±16}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±16}{2}$$

Now solve the equation x=\frac{6±16}{2} when ± is plus. Add 6 to 16.

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

Divide 22 by 2.

$$x=11$$

Now solve the equation x=\frac{6±16}{2} when ± is minus. Subtract 16 from 6.

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

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 -5 for x_{2}.

$$x^{2}-6x-55=\left(x-11\right)\left(x-\left(-5\right)\right)$$

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

$$x^{2}-6x-55=\left(x-11\right)\left(x+5\right)$$

Evaluate

$$\left(x-11\right)\left(x+5\right)$$