Factor
\left(2-x\right)\left(x-13\right)
Evaluate
\left(2-x\right)\left(x-13\right)
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a+b=15 ab=-\left(-26\right)=26
Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-26. To find a and b, set up a system to be solved.
1,26 2,13
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 26.
1+26=27 2+13=15
Calculate the sum for each pair.
a=13 b=2
The solution is the pair that gives sum 15.
\left(-x^{2}+13x\right)+\left(2x-26\right)
Rewrite -x^{2}+15x-26 as \left(-x^{2}+13x\right)+\left(2x-26\right).
-x\left(x-13\right)+2\left(x-13\right)
Factor out -x in the first and 2 in the second group.
\left(x-13\right)\left(-x+2\right)
Factor out common term x-13 by using distributive property.
-x^{2}+15x-26=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{-15±\sqrt{15^{2}-4\left(-1\right)\left(-26\right)}}{2\left(-1\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{-15±\sqrt{225-4\left(-1\right)\left(-26\right)}}{2\left(-1\right)}
Square 15.
x=\frac{-15±\sqrt{225+4\left(-26\right)}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-15±\sqrt{225-104}}{2\left(-1\right)}
Multiply 4 times -26.
x=\frac{-15±\sqrt{121}}{2\left(-1\right)}
Add 225 to -104.
x=\frac{-15±11}{2\left(-1\right)}
Take the square root of 121.
x=\frac{-15±11}{-2}
Multiply 2 times -1.
x=-\frac{4}{-2}
Now solve the equation x=\frac{-15±11}{-2} when ± is plus. Add -15 to 11.
x=2
Divide -4 by -2.
x=-\frac{26}{-2}
Now solve the equation x=\frac{-15±11}{-2} when ± is minus. Subtract 11 from -15.
x=13
Divide -26 by -2.
-x^{2}+15x-26=-\left(x-2\right)\left(x-13\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 13 for x_{2}.
x ^ 2 -15x +26 = 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 = 15 rs = 26
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{15}{2} - u s = \frac{15}{2} + u
Two numbers r and s sum up to 15 exactly when the average of the two numbers is \frac{1}{2}*15 = \frac{15}{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{15}{2} - u) (\frac{15}{2} + u) = 26
To solve for unknown quantity u, substitute these in the product equation rs = 26
\frac{225}{4} - u^2 = 26
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 26-\frac{225}{4} = -\frac{121}{4}
Simplify the expression by subtracting \frac{225}{4} on both sides
u^2 = \frac{121}{4} u = \pm\sqrt{\frac{121}{4}} = \pm \frac{11}{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{15}{2} - \frac{11}{2} = 2 s = \frac{15}{2} + \frac{11}{2} = 13
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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