Solve 2a^3+6a^2+1/2a+3/2

$$2 a ^ { 3 } + 6 a ^ { 2 } + \frac { 1 } { 2 } a + \frac { 3 } { 2 }$$

Factor

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

Solution Steps

Factor out \frac{1}{2}.

$$\frac{4a^{3}+12a^{2}+a+3}{2}$$

Consider 4a^{3}+12a^{2}+a+3. Do the grouping 4a^{3}+12a^{2}+a+3=\left(4a^{3}+12a^{2}\right)+\left(a+3\right), and factor out 4a^{2} in 4a^{3}+12a^{2}.

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

Factor out common term a+3 by using distributive property.

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

Rewrite the complete factored expression. Polynomial 4a^{2}+1 is not factored since it does not have any rational roots.

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

Evaluate

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