How do you find all real zeros of x^5 - x^4 - 3x^3 + 5x^2 - 2x = 0?

Apr 11, 2018

${x}^{5} - {x}^{4} - 3 {x}^{3} + 5 {x}^{2} - 2 x = x \left(x + 2\right) {\left(x - 1\right)}^{3}$

Explanation:

The first real zero is easy. It's zero. We can factor out an $x$.

${x}^{5} - {x}^{4} - 3 {x}^{3} + 5 {x}^{2} - 2 x = x \left({x}^{4} - {x}^{3} - 3 {x}^{2} + 5 x - 2\right)$

Now we note that when $x = - 2$,

${x}^{4} - {x}^{3} - 3 {x}^{2} + 5 x - 2$

$= {\left(- 2\right)}^{4} - {\left(- 2\right)}^{3} - 3 {\left(- 2\right)}^{2} + 5 \left(- 2\right) - 2$

$= 16 - \left(- 8\right) - 3 \left(4\right) - 10 - 2 = 0$.

This means that $x + 2$ is a factor of ${x}^{4} - {x}^{3} - 3 {x}^{2} + 5 x - 2$.

Let's factor $x + 2$ from ${x}^{4} - {x}^{3} - 3 {x}^{2} + 5 x - 2$.

${x}^{4} - {x}^{3} - 3 {x}^{2} + 5 x - 2$

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

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

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

Now we see that when $x = 1$

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

$= {1}^{3} - 3 {\left(1\right)}^{2} + 3 \left(1\right) - 1 = 1 - 3 + 3 - 1 = 0$.

This means that $x - 1$ is a factor of ${x}^{3} - 3 {x}^{2} + 3 x - 1$. Let's factor $x - 1$ from ${x}^{3} - 3 {x}^{2} + 3 x - 1$.

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

$= {x}^{3} - {x}^{2} - 2 {x}^{2} + 2 x + x - 1$

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

Finally we recognize that

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

So putting it all together, we have

${x}^{5} - {x}^{4} - 3 {x}^{3} + 5 {x}^{5} - 2 x = x \left(x + 2\right) \left(x - 1\right) \left(x - 1\right) \left(x - 1\right)$

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