Solve 6y^3-9y^2-9y+6=

$$6 y ^ { 3 } - 9 y ^ { 2 } - 9 y + 6 =$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(2y^{3}-3y^{2}-3y+2\right)$$

Consider 2y^{3}-3y^{2}-3y+2. 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 2. One such root is -1. Factor the polynomial by dividing it by y+1.

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

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

$$a+b=-5 ab=2\times 2=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

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

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

Factor out 2y in the first and -1 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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