Find roots of polynomial $$ p(x) = 49x^3-434x^2+1209x-1044 $$
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= \dfrac{ 12 }{ 7 }\\[1 em]x_3 &= \dfrac{ 29 }{ 7 } \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ 49x^3-434x^2+1209x-1044 $.
The Rational Root Theorem tells us that if the polynomial has a rational zero then it must be a fraction $ \dfrac{ \color{blue}{p}}{ \color{red}{q} } $, where $ p $ is a factor of the constant term and $ q $ is a factor of the leading coefficient.
The constant term is $ \color{blue}{ 1044 } $, with factors of 1, 2, 3, 4, 6, 9, 12, 18, 29, 36, 58, 87, 116, 174, 261, 348, 522 and 1044.
The leading coefficient is $ \color{red}{ 49 }$, with factors of 1, 7 and 49.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 1044 }}{\text{ factors of 49 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 9, 12, 18, 29, 36, 58, 87, 116, 174, 261, 348, 522, 1044 ) }}{\text{ ( 1, 7, 49 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 29}{ 1} \pm \frac{ 36}{ 1} \pm \frac{ 58}{ 1} \pm \frac{ 87}{ 1} \pm \frac{ 116}{ 1} \pm \frac{ 174}{ 1} \pm \frac{ 261}{ 1} \pm \frac{ 348}{ 1} \pm \frac{ 522}{ 1} \pm \frac{ 1044}{ 1} ~~ \pm \frac{ 1}{ 7} \pm \frac{ 2}{ 7} \pm \frac{ 3}{ 7} \pm \frac{ 4}{ 7} \pm \frac{ 6}{ 7} \pm \frac{ 9}{ 7} \pm \frac{ 12}{ 7} \pm \frac{ 18}{ 7} \pm \frac{ 29}{ 7} \pm \frac{ 36}{ 7} \pm \frac{ 58}{ 7} \pm \frac{ 87}{ 7} \pm \frac{ 116}{ 7} \pm \frac{ 174}{ 7} \pm \frac{ 261}{ 7} \pm \frac{ 348}{ 7} \pm \frac{ 522}{ 7} \pm \frac{ 1044}{ 7} ~~ \pm \frac{ 1}{ 49} \pm \frac{ 2}{ 49} \pm \frac{ 3}{ 49} \pm \frac{ 4}{ 49} \pm \frac{ 6}{ 49} \pm \frac{ 9}{ 49} \pm \frac{ 12}{ 49} \pm \frac{ 18}{ 49} \pm \frac{ 29}{ 49} \pm \frac{ 36}{ 49} \pm \frac{ 58}{ 49} \pm \frac{ 87}{ 49} \pm \frac{ 116}{ 49} \pm \frac{ 174}{ 49} \pm \frac{ 261}{ 49} \pm \frac{ 348}{ 49} \pm \frac{ 522}{ 49} \pm \frac{ 1044}{ 49} ~~ \end{aligned} $$Substitute the possible roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.
We can see that $ p\left( 3 \right) = 0 $ so $ x = 3 $ is a root of a polynomial $ p(x) $.
To find remaining zeros we use Factor Theorem. This theorem states that if $ \dfrac{p}{q} $ is root of the polynomial then the polynomial can be divided by $ \color{blue}{qx − p} $. In this example we divide polynomial $ p $ by $ \color{blue}{ x-3 }$
$$ \frac{ 49x^3-434x^2+1209x-1044}{ x-3} = 49x^2-287x+348 $$Step 2:
The next rational root is $ x = 3 $
$$ \frac{ 49x^3-434x^2+1209x-1044}{ x-3} = 49x^2-287x+348 $$Step 3:
The next rational root is $ x = \dfrac{ 12 }{ 7 } $
$$ \frac{ 49x^2-287x+348}{ 7x-12} = 7x-29 $$Step 4:
To find the last zero, solve equation $ 7x-29 = 0 $
$$ \begin{aligned} 7x-29 & = 0 \\[1 em] 7 \cdot x & = 29 \\[1 em] x & = \frac{ 29 }{ 7 } \end{aligned} $$