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x\left(3x^{2}+2x-5\right)
Factor out x.
a+b=2 ab=3\left(-5\right)=-15
Consider 3x^{2}+2x-5. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-5. To find a and b, set up a system to be solved.
-1,15 -3,5
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -15.
-1+15=14 -3+5=2
Calculate the sum for each pair.
a=-3 b=5
The solution is the pair that gives sum 2.
\left(3x^{2}-3x\right)+\left(5x-5\right)
Rewrite 3x^{2}+2x-5 as \left(3x^{2}-3x\right)+\left(5x-5\right).
3x\left(x-1\right)+5\left(x-1\right)
Factor out 3x in the first and 5 in the second group.
\left(x-1\right)\left(3x+5\right)
Factor out common term x-1 by using distributive property.
x\left(x-1\right)\left(3x+5\right)
Rewrite the complete factored expression.