Find roots of polynomial $$ p(x) = -x^3+19x^2-105x+171 $$

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

$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= 10.6458\\[1 em]x_3 &= 5.3542 \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ -x^3+19x^2-105x+171 $.

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}{ 171 } $, with a single factor of 1, 3, 9, 19, 57 and 171.

The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 171 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 3, 9, 19, 57, 171 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 19}{ 1} \pm \frac{ 57}{ 1} \pm \frac{ 171}{ 1} ~~ \end{aligned} $$

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 }$

$$ \frac{ -x^3+19x^2-105x+171}{ x-3} = -x^2+16x-57 $$

Step 2:

The next rational root is $ x = 3 $

$$ \frac{ -x^3+19x^2-105x+171}{ x-3} = -x^2+16x-57 $$

Step 3:

The solutions of $ -x^2+16x-57 = 0 $ are: $ x = 8-\sqrt{ 7 } ~ \text{and} ~ x = 8+\sqrt{ 7 }$.

You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.

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