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\left(3x+2\right)\left(6x^{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 -2 and q divides the leading coefficient 18. One such root is -\frac{2}{3}. Factor the polynomial by dividing it by 3x+2.
a+b=1 ab=6\left(-1\right)=-6
Consider 6x^{2}+x-1. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-1. To find a and b, set up a system to be solved.
-1,6 -2,3
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.
-1+6=5 -2+3=1
Calculate the sum for each pair.
a=-2 b=3
The solution is the pair that gives sum 1.
\left(6x^{2}-2x\right)+\left(3x-1\right)
Rewrite 6x^{2}+x-1 as \left(6x^{2}-2x\right)+\left(3x-1\right).
2x\left(3x-1\right)+3x-1
Factor out 2x in 6x^{2}-2x.
\left(3x-1\right)\left(2x+1\right)
Factor out common term 3x-1 by using distributive property.
\left(3x-1\right)\left(2x+1\right)\left(3x+2\right)
Rewrite the complete factored expression.