# How do you use the rational root theorem to find the roots of 10x^4 + x^3 + 7x^2 + x -3 = 0?

Nov 4, 2015

See explanation...

#### Explanation:

$f \left(x\right) = 10 {x}^{4} + {x}^{3} + 7 {x}^{2} + x - 3$

By the rational root theorem, any rational roots of $f \left(x\right) = 0$ are expressible in the form $\frac{p}{q}$, where $p , q \in \mathbb{Z}$, $q \ne 0$, $p$ and $q$ have no common factor apart from $\pm 1$, $p$ is a divisor of the constant term $- 3$ and $q$ a divisor of the coefficient, $10$, of the leading term.

So the possible rational roots are:

$\pm \frac{1}{10}$, $\pm \frac{1}{5}$, $\pm \frac{3}{10}$, $\pm \frac{1}{2}$, $\pm \frac{3}{5}$, $\pm 1$, $\pm \frac{3}{2}$, $\pm 3$

Let's try to narrow down the search a little.

$f \left(0\right) = - 3$
$f \left(1\right) = 10 + 1 + 7 + 1 - 3 = 16$

So there is a Real root in the interval $\left(0 , 1\right)$. Since $f \left(x\right)$ has only one change of sign, this will be the only positive root.

Let us calculate $f \left(\frac{3}{10}\right)$ to split our remaining positive rational possibilities:

$f \left(\frac{3}{10}\right) = 10 {\left(\frac{3}{10}\right)}^{4} + {\left(\frac{3}{10}\right)}^{3} + 7 {\left(\frac{3}{10}\right)}^{2} + \left(\frac{3}{10}\right) - 3$

$= \frac{81}{1000} + \frac{27}{1000} + \frac{63}{100} + \frac{3}{10} - 3$

$= \frac{81 + 27 + 630 + 300 - 3000}{1000}$

$= - \frac{1962}{1000} < 0$

So the remaining possible positive rational roots are $\frac{1}{2}$ and $\frac{3}{5}$.

$f \left(\frac{1}{2}\right) = 10 {\left(\frac{1}{2}\right)}^{4} + {\left(\frac{1}{2}\right)}^{3} + 7 {\left(\frac{1}{2}\right)}^{2} + \frac{1}{2} - 3$

$= \frac{5}{8} + \frac{1}{8} + \frac{7}{4} + \frac{1}{2} - 3$

$= \frac{5 + 1 + 14 + 4 - 24}{8} = 0$

So $x = \frac{1}{2}$ is a root and $\left(2 x - 1\right)$ is a factor of $f \left(x\right)$

If we separate out this factor, it's easy to spot the the remaining cubic factor will factor easily by grouping:

$10 {x}^{4} + {x}^{3} + 7 {x}^{2} + x - 3 = \left(2 x - 1\right) \left(5 {x}^{3} + 3 {x}^{2} + 5 x + 3\right)$

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

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

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

So the remaining rational root is $- \frac{3}{5}$ and the two other roots are $\pm \sqrt{- 1} = \pm i$.