Solve 6y^3+y^2-11y-6

$$6 y ^ { 3 } + y ^ { 2 } - 11 y - 6$$

Factor

$$6\left(2y-3\right)\left(y+1\right)\left(\frac{y}{2}+\frac{1}{3}\right)$$

Solution Steps

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 6. One such root is -1. Factor the polynomial by dividing it by y+1.

$$\left(y+1\right)\left(6y^{2}-5y-6\right)$$

Consider 6y^{2}-5y-6. Factor the expression by grouping. First, the expression needs to be rewritten as 6y^{2}+ay+by-6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=6\left(-6\right)=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=-9 b=4$$

Rewrite 6y^{2}-5y-6 as \left(6y^{2}-9y\right)+\left(4y-6\right).

$$\left(6y^{2}-9y\right)+\left(4y-6\right)$$

Factor out 3y in the first and 2 in the second group.

$$3y\left(2y-3\right)+2\left(2y-3\right)$$

Factor out common term 2y-3 by using distributive property.

$$\left(2y-3\right)\left(3y+2\right)$$

Rewrite the complete factored expression.

$$\left(2y-3\right)\left(y+1\right)\left(3y+2\right)$$

Evaluate

$$6y^{3}+y^{2}-11y-6$$