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y\left(x^{2}-3x-10\right)
Factor out y.
a+b=-3 ab=1\left(-10\right)=-10
Consider x^{2}-3x-10. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-10. To find a and b, set up a system to be solved.
1,-10 2,-5
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.
1-10=-9 2-5=-3
Calculate the sum for each pair.
a=-5 b=2
The solution is the pair that gives sum -3.
\left(x^{2}-5x\right)+\left(2x-10\right)
Rewrite x^{2}-3x-10 as \left(x^{2}-5x\right)+\left(2x-10\right).
x\left(x-5\right)+2\left(x-5\right)
Factor out x in the first and 2 in the second group.
\left(x-5\right)\left(x+2\right)
Factor out common term x-5 by using distributive property.
y\left(x-5\right)\left(x+2\right)
Rewrite the complete factored expression.