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\left(2x-3\right)\left(2x^{2}-x-1\right)
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 3 and q divides the leading coefficient 4. One such root is \frac{3}{2}. Factor the polynomial by dividing it by 2x-3.
a+b=-1 ab=2\left(-1\right)=-2
Consider 2x^{2}-x-1. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-1. To find a and b, set up a system to be solved.
a=-2 b=1
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. The only such pair is the system solution.
\left(2x^{2}-2x\right)+\left(x-1\right)
Rewrite 2x^{2}-x-1 as \left(2x^{2}-2x\right)+\left(x-1\right).
2x\left(x-1\right)+x-1
Factor out 2x in 2x^{2}-2x.
\left(x-1\right)\left(2x+1\right)
Factor out common term x-1 by using distributive property.
\left(2x-3\right)\left(x-1\right)\left(2x+1\right)
Rewrite the complete factored expression.