Solve x^3/5-2x^2/5-x+2

$$\frac{ { x }^{ 3 } }{ 5 } - \frac{ 2 { x }^{ 2 } }{ 5 } -x+2$$

Evaluate

$$\frac{\left(x-2\right)\left(x^{2}-5\right)}{5}$$

Short Solution Steps

Since \frac{x^{3}}{5} and \frac{2x^{2}}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{3}-2x^{2}}{5}-x+2$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -x+2 times \frac{5}{5}.

$$\frac{x^{3}-2x^{2}}{5}+\frac{5\left(-x+2\right)}{5}$$

Since \frac{x^{3}-2x^{2}}{5} and \frac{5\left(-x+2\right)}{5} have the same denominator, add them by adding their numerators.

$$\frac{x^{3}-2x^{2}+5\left(-x+2\right)}{5}$$

Do the multiplications in x^{3}-2x^{2}+5\left(-x+2\right).

$$\frac{x^{3}-2x^{2}-5x+10}{5}$$

Factor

$$\frac{\left(x-2\right)\left(x^{2}-5\right)}{5}$$

Solution Steps

Factor out \frac{1}{5}.

$$\frac{x^{3}-2x^{2}-5x+10}{5}$$

Consider x^{3}-2x^{2}-5x+10. Do the grouping x^{3}-2x^{2}-5x+10=\left(x^{3}-2x^{2}\right)+\left(-5x+10\right), and factor out x^{2} in the first and -5 in the second group.

$$x^{2}\left(x-2\right)-5\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x^{2}-5\right)$$

Rewrite the complete factored expression. Polynomial x^{2}-5 is not factored since it does not have any rational roots.

$$\frac{\left(x-2\right)\left(x^{2}-5\right)}{5}$$