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a+b=15 ab=1\times 50=50
Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+50. To find a and b, set up a system to be solved.
1,50 2,25 5,10
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 50.
1+50=51 2+25=27 5+10=15
Calculate the sum for each pair.
a=5 b=10
The solution is the pair that gives sum 15.
\left(x^{2}+5x\right)+\left(10x+50\right)
Rewrite x^{2}+15x+50 as \left(x^{2}+5x\right)+\left(10x+50\right).
x\left(x+5\right)+10\left(x+5\right)
Factor out x in the first and 10 in the second group.
\left(x+5\right)\left(x+10\right)
Factor out common term x+5 by using distributive property.
x^{2}+15x+50=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\times 50}}{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{-15±\sqrt{225-4\times 50}}{2}
Square 15.
x=\frac{-15±\sqrt{225-200}}{2}
Multiply -4 times 50.
x=\frac{-15±\sqrt{25}}{2}
Add 225 to -200.
x=\frac{-15±5}{2}
Take the square root of 25.
x=-\frac{10}{2}
Now solve the equation x=\frac{-15±5}{2} when ± is plus. Add -15 to 5.
x=-5
Divide -10 by 2.
x=-\frac{20}{2}
Now solve the equation x=\frac{-15±5}{2} when ± is minus. Subtract 5 from -15.
x=-10
Divide -20 by 2.
x^{2}+15x+50=\left(x-\left(-5\right)\right)\left(x-\left(-10\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5 for x_{1} and -10 for x_{2}.
x^{2}+15x+50=\left(x+5\right)\left(x+10\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.