Solve 6x^3+8x^2+2x

$$6 { x }^{ 3 } +8 { x }^{ 2 } +2x$$

Factor

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

Solution Steps

Factor out 2.

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

Consider 3x^{3}+4x^{2}+x. Factor out x.

$$x\left(3x^{2}+4x+1\right)$$

Consider 3x^{2}+4x+1. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=4 ab=3\times 1=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=3$$

Rewrite 3x^{2}+4x+1 as \left(3x^{2}+x\right)+\left(3x+1\right).

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

Factor out x in 3x^{2}+x.

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

Factor out common term 3x+1 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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