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a+b=-1 ab=1\left(-2\right)=-2
Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-2. To find a and b, set up a system to be solved.
a=-2 b=1
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. The only such pair is the system solution.
\left(x^{2}-2x\right)+\left(x-2\right)
Rewrite x^{2}-x-2 as \left(x^{2}-2x\right)+\left(x-2\right).
x\left(x-2\right)+x-2
Factor out x in x^{2}-2x.
\left(x-2\right)\left(x+1\right)
Factor out common term x-2 by using distributive property.
x^{2}-x-2=0
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=\frac{-\left(-1\right)±\sqrt{1-4\left(-2\right)}}{2}
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+8}}{2}
Multiply -4 times -2.
x=\frac{-\left(-1\right)±\sqrt{9}}{2}
Add 1 to 8.
x=\frac{-\left(-1\right)±3}{2}
Take the square root of 9.
x=\frac{1±3}{2}
The opposite of -1 is 1.
x=\frac{4}{2}
Now solve the equation x=\frac{1±3}{2} when ± is plus. Add 1 to 3.
x=2
Divide 4 by 2.
x=-\frac{2}{2}
Now solve the equation x=\frac{1±3}{2} when ± is minus. Subtract 3 from 1.
x=-1
Divide -2 by 2.
x^{2}-x-2=\left(x-2\right)\left(x-\left(-1\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -1 for x_{2}.
x^{2}-x-2=\left(x-2\right)\left(x+1\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.