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-x^{2}+5x-6
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=5 ab=-\left(-6\right)=6
Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-6. To find a and b, set up a system to be solved.
1,6 2,3
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 6.
1+6=7 2+3=5
Calculate the sum for each pair.
a=3 b=2
The solution is the pair that gives sum 5.
\left(-x^{2}+3x\right)+\left(2x-6\right)
Rewrite -x^{2}+5x-6 as \left(-x^{2}+3x\right)+\left(2x-6\right).
-x\left(x-3\right)+2\left(x-3\right)
Factor out -x in the first and 2 in the second group.
\left(x-3\right)\left(-x+2\right)
Factor out common term x-3 by using distributive property.
-x^{2}+5x-6=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{-5±\sqrt{5^{2}-4\left(-1\right)\left(-6\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{-5±\sqrt{25-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}
Square 5.
x=\frac{-5±\sqrt{25+4\left(-6\right)}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-5±\sqrt{25-24}}{2\left(-1\right)}
Multiply 4 times -6.
x=\frac{-5±\sqrt{1}}{2\left(-1\right)}
Add 25 to -24.
x=\frac{-5±1}{2\left(-1\right)}
Take the square root of 1.
x=\frac{-5±1}{-2}
Multiply 2 times -1.
x=-\frac{4}{-2}
Now solve the equation x=\frac{-5±1}{-2} when ± is plus. Add -5 to 1.
x=2
Divide -4 by -2.
x=-\frac{6}{-2}
Now solve the equation x=\frac{-5±1}{-2} when ± is minus. Subtract 1 from -5.
x=3
Divide -6 by -2.
-x^{2}+5x-6=-\left(x-2\right)\left(x-3\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 3 for x_{2}.