Solve n^2-n-110

$$n ^ { 2 } - n - 110$$

Factor

$$\left(n-11\right)\left(n+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as n^{2}+an+bn-110. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-110\right)=-110$$

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 -110.

$$1,-110 2,-55 5,-22 10,-11$$

Calculate the sum for each pair.

$$1-110=-109 2-55=-53 5-22=-17 10-11=-1$$

The solution is the pair that gives sum -1.

$$a=-11 b=10$$

Rewrite n^{2}-n-110 as \left(n^{2}-11n\right)+\left(10n-110\right).

$$\left(n^{2}-11n\right)+\left(10n-110\right)$$

Factor out n in the first and 10 in the second group.

$$n\left(n-11\right)+10\left(n-11\right)$$

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

$$\left(n-11\right)\left(n+10\right)$$

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.

$$n^{2}-n-110=0$$

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.

$$n=\frac{-\left(-1\right)±\sqrt{1-4\left(-110\right)}}{2}$$

Multiply -4 times -110.

$$n=\frac{-\left(-1\right)±\sqrt{1+440}}{2}$$

Add 1 to 440.

$$n=\frac{-\left(-1\right)±\sqrt{441}}{2}$$

Take the square root of 441.

$$n=\frac{-\left(-1\right)±21}{2}$$

The opposite of -1 is 1.

$$n=\frac{1±21}{2}$$

Now solve the equation n=\frac{1±21}{2} when ± is plus. Add 1 to 21.

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

Divide 22 by 2.

$$n=11$$

Now solve the equation n=\frac{1±21}{2} when ± is minus. Subtract 21 from 1.

$$n=-\frac{20}{2}$$

Divide -20 by 2.

$$n=-10$$

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 -10 for x_{2}.

$$n^{2}-n-110=\left(n-11\right)\left(n-\left(-10\right)\right)$$

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

$$n^{2}-n-110=\left(n-11\right)\left(n+10\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -1x -110 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 1 rs = -110$$

Two numbers r and s sum up to 1 exactly when the average of the two numbers is \frac{1}{2}*1 = \frac{1}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{1}{2} - u s = \frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -110

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -110$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{1}{4} - u^2 = -110$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -110-\frac{1}{4} = -\frac{441}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{441}{4} u = \pm\sqrt{\frac{441}{4}} = \pm \frac{21}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{1}{2} - \frac{21}{2} = -10 s = \frac{1}{2} + \frac{21}{2} = 11$$

Evaluate

$$\left(n-11\right)\left(n+10\right)$$