# How do you use the rational root theorem to find the roots of x^3-9x^2+14x=48?

Feb 3, 2016

Test all possible values of $x \in \left\{\pm \left(\text{factors of " 48)/("factors of } 1\right)\right\}$
Any that satisfy the equation are roots of the equation.
$\textcolor{w h i t e}{\text{XXX}}$(The only Real root is at $x = 8$ in this case).

#### Explanation:

The candidates for rational roots of $\textcolor{red}{1} {x}^{3} - 9 {x}^{2} + 14 x - \textcolor{b l u e}{48} = 0$
(notice, I've slightly re-arranged the equation)
are $\pm \left(\text{factors of "color(blue)(48))/("factors of } \textcolor{red}{1}\right)$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXX}} = \pm \left(1 , 2 , 3 , 4 , 6 , 8 , 12 , 16 , 24 , 48\right)$

Therefore there are $20$ candidates to test!
(It is useful if you can set this up as a spreadsheet or computer program.)

Here are the evaluations (using synthetic substitution/division) for the first $8$ candidates:

This gives us $x = 8$ as a zero of the given equation.

Not of the subsequent factors gave a result of $0$ (Take my word for it).

It might be noted that treating these entries as synthetic division gives:
$\textcolor{w h i t e}{\text{XXX}} {x}^{3} - 9 {x}^{2} + 14 x - 48 = \left(x - 8\right) \left({x}^{2} - x + 6\right)$
and taking the discriminant of $\left({x}^{2} - x + 6\right)$ shows us that there are no further Real roots (i.e. no other Real zeros)