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\frac{4x^{2}-x-3}{3}
Factor out \frac{1}{3}.
a+b=-1 ab=4\left(-3\right)=-12
Consider 4x^{2}-x-3. Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-3. To find a and b, set up a system to be solved.
1,-12 2,-6 3,-4
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 -12.
1-12=-11 2-6=-4 3-4=-1
Calculate the sum for each pair.
a=-4 b=3
The solution is the pair that gives sum -1.
\left(4x^{2}-4x\right)+\left(3x-3\right)
Rewrite 4x^{2}-x-3 as \left(4x^{2}-4x\right)+\left(3x-3\right).
4x\left(x-1\right)+3\left(x-1\right)
Factor out 4x in the first and 3 in the second group.
\left(x-1\right)\left(4x+3\right)
Factor out common term x-1 by using distributive property.
\frac{\left(x-1\right)\left(4x+3\right)}{3}
Rewrite the complete factored expression.