Solve -x/3+1/3+(2x^2-x/3-2/3)(3x^2+2)

$$- x / 3 + 1 / 3 + ( 2 x ^ { 2 } - x / 3 - 2 / 3 ) ( 3 x ^ { 2 } + 2 )$$

Evaluate

$$6x^{4}-x^{3}+2x^{2}-x-1$$

Short Solution Steps

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

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

Since \frac{-x}{3} and \frac{1}{3} have the same denominator, add them by adding their numerators.

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

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

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

Since \frac{3\times 2x^{2}}{3} and \frac{-x-2}{3} have the same denominator, add them by adding their numerators.

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

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

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

Express \frac{6x^{2}-x-2}{3}\left(3x^{2}+2\right) as a single fraction.

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

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

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

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

$$\frac{-x+1+18x^{4}+12x^{2}-3x^{3}-2x-6x^{2}-4}{3}$$

Combine like terms in -x+1+18x^{4}+12x^{2}-3x^{3}-2x-6x^{2}-4.

$$\frac{-3x-3+18x^{4}+6x^{2}-3x^{3}}{3}$$

Divide each term of -3x-3+18x^{4}+6x^{2}-3x^{3} by 3 to get -x-1+6x^{4}+2x^{2}-x^{3}.

$$-x-1+6x^{4}+2x^{2}-x^{3}$$

Expand

$$6x^{4}-x^{3}+2x^{2}-x-1$$

Short Solution Steps

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

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

Since \frac{-x}{3} and \frac{1}{3} have the same denominator, add them by adding their numerators.

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

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

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

Since \frac{3\times 2x^{2}}{3} and \frac{-x-2}{3} have the same denominator, add them by adding their numerators.

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

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

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

Express \frac{6x^{2}-x-2}{3}\left(3x^{2}+2\right) as a single fraction.

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

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

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

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

$$\frac{-x+1+18x^{4}+12x^{2}-3x^{3}-2x-6x^{2}-4}{3}$$

Combine like terms in -x+1+18x^{4}+12x^{2}-3x^{3}-2x-6x^{2}-4.

$$\frac{-3x-3+18x^{4}+6x^{2}-3x^{3}}{3}$$

Divide each term of -3x-3+18x^{4}+6x^{2}-3x^{3} by 3 to get -x-1+6x^{4}+2x^{2}-x^{3}.

$$-x-1+6x^{4}+2x^{2}-x^{3}$$