Solve =[x^3/3-3x^2/2+2x+c]

$$= [ \frac { x ^ { 3 } } { 3 } - \frac { 3 x ^ { 2 } } { 2 } + 2 x + c ]$$

Evaluate

$$\frac{x^{3}}{3}-\frac{3x^{2}}{2}+c+2x$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 2 is 6. Multiply \frac{x^{3}}{3} times \frac{2}{2}. Multiply \frac{3x^{2}}{2} times \frac{3}{3}.

$$\frac{2x^{3}}{6}-\frac{3\times 3x^{2}}{6}+2x+c$$

Since \frac{2x^{3}}{6} and \frac{3\times 3x^{2}}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x^{3}-3\times 3x^{2}}{6}+2x+c$$

Do the multiplications in 2x^{3}-3\times 3x^{2}.

$$\frac{2x^{3}-9x^{2}}{6}+2x+c$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x+c times \frac{6}{6}.

$$\frac{2x^{3}-9x^{2}}{6}+\frac{6\left(2x+c\right)}{6}$$

Since \frac{2x^{3}-9x^{2}}{6} and \frac{6\left(2x+c\right)}{6} have the same denominator, add them by adding their numerators.

$$\frac{2x^{3}-9x^{2}+6\left(2x+c\right)}{6}$$

Do the multiplications in 2x^{3}-9x^{2}+6\left(2x+c\right).

$$\frac{2x^{3}-9x^{2}+12x+6c}{6}$$

Factor

$$\frac{2x^{3}-9x^{2}+12x+6c}{6}$$

Solution Steps

Factor out \frac{1}{6}.

$$\frac{2x^{3}-9x^{2}+12x+6c}{6}$$