Solve 6x^4+19x^3+11x^2-6x

$$6 x ^ { 4 } + 19 x ^ { 3 } + 11 x ^ { 2 } - 6 x$$

Factor

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

Solution Steps

Factor out x.

$$x\left(6x^{3}+19x^{2}+11x-6\right)$$

Consider 6x^{3}+19x^{2}+11x-6. 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 \frac{1}{3}. Factor the polynomial by dividing it by 3x-1.

$$\left(3x-1\right)\left(2x^{2}+7x+6\right)$$

Consider 2x^{2}+7x+6. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=7 ab=2\times 6=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite 2x^{2}+7x+6 as \left(2x^{2}+3x\right)+\left(4x+6\right).

$$\left(2x^{2}+3x\right)+\left(4x+6\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(2x+3\right)+2\left(2x+3\right)$$

Factor out common term 2x+3 by using distributive property.

$$\left(2x+3\right)\left(x+2\right)$$

Rewrite the complete factored expression.

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

Evaluate

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