Factor
\left(1-x\right)\left(2x-7\right)
Evaluate
\left(1-x\right)\left(2x-7\right)
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a+b=9 ab=-2\left(-7\right)=14
Factor the expression by grouping. First, the expression needs to be rewritten as -2x^{2}+ax+bx-7. To find a and b, set up a system to be solved.
1,14 2,7
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 14.
1+14=15 2+7=9
Calculate the sum for each pair.
a=7 b=2
The solution is the pair that gives sum 9.
\left(-2x^{2}+7x\right)+\left(2x-7\right)
Rewrite -2x^{2}+9x-7 as \left(-2x^{2}+7x\right)+\left(2x-7\right).
-x\left(2x-7\right)+2x-7
Factor out -x in -2x^{2}+7x.
\left(2x-7\right)\left(-x+1\right)
Factor out common term 2x-7 by using distributive property.
-2x^{2}+9x-7=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{-9±\sqrt{9^{2}-4\left(-2\right)\left(-7\right)}}{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{-9±\sqrt{81-4\left(-2\right)\left(-7\right)}}{2\left(-2\right)}
Square 9.
x=\frac{-9±\sqrt{81+8\left(-7\right)}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-9±\sqrt{81-56}}{2\left(-2\right)}
Multiply 8 times -7.
x=\frac{-9±\sqrt{25}}{2\left(-2\right)}
Add 81 to -56.
x=\frac{-9±5}{2\left(-2\right)}
Take the square root of 25.
x=\frac{-9±5}{-4}
Multiply 2 times -2.
x=-\frac{4}{-4}
Now solve the equation x=\frac{-9±5}{-4} when ± is plus. Add -9 to 5.
x=1
Divide -4 by -4.
x=-\frac{14}{-4}
Now solve the equation x=\frac{-9±5}{-4} when ± is minus. Subtract 5 from -9.
x=\frac{7}{2}
Reduce the fraction \frac{-14}{-4} to lowest terms by extracting and canceling out 2.
-2x^{2}+9x-7=-2\left(x-1\right)\left(x-\frac{7}{2}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and \frac{7}{2} for x_{2}.
-2x^{2}+9x-7=-2\left(x-1\right)\times \frac{-2x+7}{-2}
Subtract \frac{7}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
-2x^{2}+9x-7=\left(x-1\right)\left(-2x+7\right)
Cancel out 2, the greatest common factor in -2 and 2.
x ^ 2 -\frac{9}{2}x +\frac{7}{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 = \frac{9}{2} rs = \frac{7}{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{9}{4} - u s = \frac{9}{4} + u
Two numbers r and s sum up to \frac{9}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{2} = \frac{9}{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{9}{4} - u) (\frac{9}{4} + u) = \frac{7}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{2}
\frac{81}{16} - u^2 = \frac{7}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{7}{2}-\frac{81}{16} = -\frac{25}{16}
Simplify the expression by subtracting \frac{81}{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{9}{4} - \frac{5}{4} = 1 s = \frac{9}{4} + \frac{5}{4} = 3.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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