Find roots of polynomial $$ p(x) = 245x^3-161x^2-17x+5 $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= -\frac{ 1 }{ 5 }\\[1 em]x_2 &= \frac{ 1 }{ 7 }\\[1 em]x_3 &= \frac{ 5 }{ 7 } \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = -\dfrac{ 1 }{ 5 } } $ is a root of polynomial $ 245x^3-161x^2-17x+5 $.

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}{ 5 } $, with factors of 1 and 5.

The leading coefficient is $ \color{red}{ 245 }$, with factors of 1, 5, 7, 35, 49 and 245.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 5 }}{\text{ factors of 245 }} = \pm \dfrac{\text{ ( 1, 5 ) }}{\text{ ( 1, 5, 7, 35, 49, 245 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 5}{ 1} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 5}{ 5} ~~ \pm \frac{ 1}{ 7} \pm \frac{ 5}{ 7} ~~ \pm \frac{ 1}{ 35} \pm \frac{ 5}{ 35} ~~ \pm \frac{ 1}{ 49} \pm \frac{ 5}{ 49} ~~ \pm \frac{ 1}{ 245} \pm \frac{ 5}{ 245} ~~ \end{aligned} $$

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( -\dfrac{ 1 }{ 5 } \right) = 0 $ so $ x = -\dfrac{ 1 }{ 5 } $ 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}{ 5x+1 }$

$$ \frac{ 245x^3-161x^2-17x+5}{ 5x+1} = 49x^2-42x+5 $$

Step 2:

The next rational root is $ x = -\dfrac{ 1 }{ 5 } $

$$ \frac{ 245x^3-161x^2-17x+5}{ 5x+1} = 49x^2-42x+5 $$

Step 3:

The next rational root is $ x = \dfrac{ 1 }{ 7 } $

$$ \frac{ 49x^2-42x+5}{ 7x-1} = 7x-5 $$

Step 4:

To find the last zero, solve equation $ 7x-5 = 0 $

$$ \begin{aligned} 7x-5 & = 0 \\[1 em] 7 \cdot x & = 5 \\[1 em] x & = \frac{ 5 }{ 7 } \end{aligned} $$

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