# What are the possible rational roots of x^4-5x^3+9x^2-7x+2=0 and then determine the rational roots?

Feb 4, 2017

The "possible" rational roots are: $\pm 1$, $\pm 2$

The actual roots are: $1$, $1$, $1$, $2$.

#### Explanation:

Given:

$f \left(x\right) = {x}^{4} - 5 {x}^{3} + 9 {x}^{2} - 7 x + 2$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $2$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2$

Note also that the pattern of signs of the coefficients is: $+ - + - +$. With $4$ changes, we can use Descartes' Rule of Signs to deduce that means that $f \left(x\right)$ has $4$, $2$ or $0$ positive Real zeros. The signs of the coefficients of $f \left(- x\right)$ are $+ + + + +$, meaning that $f \left(x\right)$ has no negative zeros.

So the only possible rational zeros are $1$ or $2$.

We find:

$f \left(1\right) = 1 - 5 + 9 - 7 + 2 = 0$

So $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{4} - 5 {x}^{3} + 9 {x}^{2} - 7 x + 2 = \left(x - 1\right) \left({x}^{3} - 4 {x}^{2} + 5 x - 2\right)$

Note that the sum of the coefficients of the remaining cubic factor is also zero, so $x = 1$ is a zero again and $\left(x - 1\right)$ a factor:

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

Note that the remaining quadratic also has $x = 1$ as a zero and $\left(x - 1\right)$ as a factor:

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