Factor
\left(x-1\right)\left(2x-5\right)
Evaluate
\left(x-1\right)\left(2x-5\right)
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a+b=-7 ab=2\times 5=10
Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+5. To find a and b, set up a system to be solved.
-1,-10 -2,-5
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 10.
-1-10=-11 -2-5=-7
Calculate the sum for each pair.
a=-5 b=-2
The solution is the pair that gives sum -7.
\left(2x^{2}-5x\right)+\left(-2x+5\right)
Rewrite 2x^{2}-7x+5 as \left(2x^{2}-5x\right)+\left(-2x+5\right).
x\left(2x-5\right)-\left(2x-5\right)
Factor out x in the first and -1 in the second group.
\left(2x-5\right)\left(x-1\right)
Factor out common term 2x-5 by using distributive property.
2x^{2}-7x+5=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(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 5}}{2\times 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(-7\right)±\sqrt{49-4\times 2\times 5}}{2\times 2}
Square -7.
x=\frac{-\left(-7\right)±\sqrt{49-8\times 5}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-7\right)±\sqrt{49-40}}{2\times 2}
Multiply -8 times 5.
x=\frac{-\left(-7\right)±\sqrt{9}}{2\times 2}
Add 49 to -40.
x=\frac{-\left(-7\right)±3}{2\times 2}
Take the square root of 9.
x=\frac{7±3}{2\times 2}
The opposite of -7 is 7.
x=\frac{7±3}{4}
Multiply 2 times 2.
x=\frac{10}{4}
Now solve the equation x=\frac{7±3}{4} when ± is plus. Add 7 to 3.
x=\frac{5}{2}
Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.
x=\frac{4}{4}
Now solve the equation x=\frac{7±3}{4} when ± is minus. Subtract 3 from 7.
x=1
Divide 4 by 4.
2x^{2}-7x+5=2\left(x-\frac{5}{2}\right)\left(x-1\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and 1 for x_{2}.
2x^{2}-7x+5=2\times \frac{2x-5}{2}\left(x-1\right)
Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
2x^{2}-7x+5=\left(2x-5\right)\left(x-1\right)
Cancel out 2, the greatest common factor in 2 and 2.
x ^ 2 -\frac{7}{2}x +\frac{5}{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.This is achieved by dividing both sides of the equation by 2
r + s = \frac{7}{2} rs = \frac{5}{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{7}{4} - u s = \frac{7}{4} + u
Two numbers r and s sum up to \frac{7}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{2} = \frac{7}{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{7}{4} - u) (\frac{7}{4} + u) = \frac{5}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{2}
\frac{49}{16} - u^2 = \frac{5}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{5}{2}-\frac{49}{16} = -\frac{9}{16}
Simplify the expression by subtracting \frac{49}{16} on both sides
u^2 = \frac{9}{16} u = \pm\sqrt{\frac{9}{16}} = \pm \frac{3}{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{7}{4} - \frac{3}{4} = 1 s = \frac{7}{4} + \frac{3}{4} = 2.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
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Linear equation
y = 3x + 4
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Matrix
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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
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