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3\left(2a^{2}b+3ab-2b\right)
Factor out 3.
b\left(2a^{2}+3a-2\right)
Consider 2a^{2}b+3ab-2b. Factor out b.
p+q=3 pq=2\left(-2\right)=-4
Consider 2a^{2}+3a-2. Factor the expression by grouping. First, the expression needs to be rewritten as 2a^{2}+pa+qa-2. To find p and q, set up a system to be solved.
-1,4 -2,2
Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.
-1+4=3 -2+2=0
Calculate the sum for each pair.
p=-1 q=4
The solution is the pair that gives sum 3.
\left(2a^{2}-a\right)+\left(4a-2\right)
Rewrite 2a^{2}+3a-2 as \left(2a^{2}-a\right)+\left(4a-2\right).
a\left(2a-1\right)+2\left(2a-1\right)
Factor out a in the first and 2 in the second group.
\left(2a-1\right)\left(a+2\right)
Factor out common term 2a-1 by using distributive property.
3b\left(2a-1\right)\left(a+2\right)
Rewrite the complete factored expression.