Factor
\left(x-2\right)\left(x+1\right)
Evaluate
\left(x-2\right)\left(x+1\right)
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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.
x ^ 2 -1x -2 = 0
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.
r + s = 1 rs = -2
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 = \frac{1}{2} - u s = \frac{1}{2} + u
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>
(\frac{1}{2} - u) (\frac{1}{2} + u) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
\frac{1}{4} - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-\frac{1}{4} = -\frac{9}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{2} - \frac{3}{2} = -1 s = \frac{1}{2} + \frac{3}{2} = 2
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}