Factor
\left(2-x\right)\left(2x+1\right)
Evaluate
\left(2-x\right)\left(2x+1\right)
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a+b=3 ab=-2\times 2=-4
Factor the expression by grouping. First, the expression needs to be rewritten as -2x^{2}+ax+bx+2. To find a and b, set up a system to be solved.
-1,4 -2,2
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.
-1+4=3 -2+2=0
Calculate the sum for each pair.
a=4 b=-1
The solution is the pair that gives sum 3.
\left(-2x^{2}+4x\right)+\left(-x+2\right)
Rewrite -2x^{2}+3x+2 as \left(-2x^{2}+4x\right)+\left(-x+2\right).
2x\left(-x+2\right)-x+2
Factor out 2x in -2x^{2}+4x.
\left(-x+2\right)\left(2x+1\right)
Factor out common term -x+2 by using distributive property.
-2x^{2}+3x+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{-3±\sqrt{3^{2}-4\left(-2\right)\times 2}}{2\left(-2\right)}
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{-3±\sqrt{9-4\left(-2\right)\times 2}}{2\left(-2\right)}
Square 3.
x=\frac{-3±\sqrt{9+8\times 2}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-3±\sqrt{9+16}}{2\left(-2\right)}
Multiply 8 times 2.
x=\frac{-3±\sqrt{25}}{2\left(-2\right)}
Add 9 to 16.
x=\frac{-3±5}{2\left(-2\right)}
Take the square root of 25.
x=\frac{-3±5}{-4}
Multiply 2 times -2.
x=\frac{2}{-4}
Now solve the equation x=\frac{-3±5}{-4} when ± is plus. Add -3 to 5.
x=-\frac{1}{2}
Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.
x=-\frac{8}{-4}
Now solve the equation x=\frac{-3±5}{-4} when ± is minus. Subtract 5 from -3.
x=2
Divide -8 by -4.
-2x^{2}+3x+2=-2\left(x-\left(-\frac{1}{2}\right)\right)\left(x-2\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{2} for x_{1} and 2 for x_{2}.
-2x^{2}+3x+2=-2\left(x+\frac{1}{2}\right)\left(x-2\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
-2x^{2}+3x+2=-2\times \frac{-2x-1}{-2}\left(x-2\right)
Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
-2x^{2}+3x+2=\left(-2x-1\right)\left(x-2\right)
Cancel out 2, the greatest common factor in -2 and 2.
x ^ 2 -\frac{3}{2}x -1 = 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 = \frac{3}{2} rs = -1
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{3}{4} - u s = \frac{3}{4} + u
Two numbers r and s sum up to \frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{2} = \frac{3}{4}. 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{3}{4} - u) (\frac{3}{4} + u) = -1
To solve for unknown quantity u, substitute these in the product equation rs = -1
\frac{9}{16} - u^2 = -1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -1-\frac{9}{16} = -\frac{25}{16}
Simplify the expression by subtracting \frac{9}{16} on both sides
u^2 = \frac{25}{16} u = \pm\sqrt{\frac{25}{16}} = \pm \frac{5}{4}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{3}{4} - \frac{5}{4} = -0.500 s = \frac{3}{4} + \frac{5}{4} = 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}