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\left(x-3\right)\left(2x^{2}-3x-2\right)
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 2. One such root is 3. Factor the polynomial by dividing it by x-3.
a+b=-3 ab=2\left(-2\right)=-4
Consider 2x^{2}-3x-2. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-2. To find a and b, set up a system to be solved.
1,-4 2,-2
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 -4.
1-4=-3 2-2=0
Calculate the sum for each pair.
a=-4 b=1
The solution is the pair that gives sum -3.
\left(2x^{2}-4x\right)+\left(x-2\right)
Rewrite 2x^{2}-3x-2 as \left(2x^{2}-4x\right)+\left(x-2\right).
2x\left(x-2\right)+x-2
Factor out 2x in 2x^{2}-4x.
\left(x-2\right)\left(2x+1\right)
Factor out common term x-2 by using distributive property.
\left(x-3\right)\left(x-2\right)\left(2x+1\right)
Rewrite the complete factored expression.