Solve x^2-x-210

$${ x }^{ 2 } -x-210$$

Factor

$$\left(x-15\right)\left(x+14\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-210. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-210\right)=-210$$

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 -210.

$$1,-210 2,-105 3,-70 5,-42 6,-35 7,-30 10,-21 14,-15$$

Calculate the sum for each pair.

$$1-210=-209 2-105=-103 3-70=-67 5-42=-37 6-35=-29 7-30=-23 10-21=-11 14-15=-1$$

The solution is the pair that gives sum -1.

$$a=-15 b=14$$

Rewrite x^{2}-x-210 as \left(x^{2}-15x\right)+\left(14x-210\right).

$$\left(x^{2}-15x\right)+\left(14x-210\right)$$

Factor out x in the first and 14 in the second group.

$$x\left(x-15\right)+14\left(x-15\right)$$

Factor out common term x-15 by using distributive property.

$$\left(x-15\right)\left(x+14\right)$$

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.

$$x^{2}-x-210=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-210\right)}}{2}$$

Multiply -4 times -210.

$$x=\frac{-\left(-1\right)±\sqrt{1+840}}{2}$$

Add 1 to 840.

$$x=\frac{-\left(-1\right)±\sqrt{841}}{2}$$

Take the square root of 841.

$$x=\frac{-\left(-1\right)±29}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±29}{2}$$

Now solve the equation x=\frac{1±29}{2} when ± is plus. Add 1 to 29.

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

Divide 30 by 2.

$$x=15$$

Now solve the equation x=\frac{1±29}{2} when ± is minus. Subtract 29 from 1.

$$x=-\frac{28}{2}$$

Divide -28 by 2.

$$x=-14$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 15 for x_{1} and -14 for x_{2}.

$$x^{2}-x-210=\left(x-15\right)\left(x-\left(-14\right)\right)$$

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

$$x^{2}-x-210=\left(x-15\right)\left(x+14\right)$$

Evaluate

$$\left(x-15\right)\left(x+14\right)$$