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5\left(2a^{2}-a+3a^{3}\right)
Factor out 5.
a\left(2a-1+3a^{2}\right)
Consider 2a^{2}-a+3a^{3}. Factor out a.
3a^{2}+2a-1
Consider 2a-1+3a^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
p+q=2 pq=3\left(-1\right)=-3
Factor the expression by grouping. First, the expression needs to be rewritten as 3a^{2}+pa+qa-1. To find p and q, set up a system to be solved.
p=-1 q=3
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. The only such pair is the system solution.
\left(3a^{2}-a\right)+\left(3a-1\right)
Rewrite 3a^{2}+2a-1 as \left(3a^{2}-a\right)+\left(3a-1\right).
a\left(3a-1\right)+3a-1
Factor out a in 3a^{2}-a.
\left(3a-1\right)\left(a+1\right)
Factor out common term 3a-1 by using distributive property.
5a\left(3a-1\right)\left(a+1\right)
Rewrite the complete factored expression.