Solve 9x^3+6x^2+xdiv3x+1

$$9 { x }^{ 3 } +6 { x }^{ 2 } +x \div 3x+1$$

Evaluate

$$9x^{3}+\frac{19x^{2}}{3}+1$$

Short Solution Steps

Express \frac{x}{3}x as a single fraction.

$$9x^{3}+6x^{2}+\frac{xx}{3}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9x^{3}+6x^{2}+1 times \frac{3}{3}.

$$\frac{3\left(9x^{3}+6x^{2}+1\right)}{3}+\frac{xx}{3}$$

Since \frac{3\left(9x^{3}+6x^{2}+1\right)}{3} and \frac{xx}{3} have the same denominator, add them by adding their numerators.

$$\frac{3\left(9x^{3}+6x^{2}+1\right)+xx}{3}$$

Do the multiplications in 3\left(9x^{3}+6x^{2}+1\right)+xx.

$$\frac{27x^{3}+18x^{2}+3+x^{2}}{3}$$

Combine like terms in 27x^{3}+18x^{2}+3+x^{2}.

$$\frac{27x^{3}+19x^{2}+3}{3}$$

Factor

$$\frac{27x^{3}+19x^{2}+3}{3}$$

Solution Steps

Factor out \frac{1}{3}. Polynomial 27x^{3}+19x^{2}+3 is not factored since it does not have any rational roots.

$$\frac{27x^{3}+18x^{2}+xx+3}{3}$$