Find roots of polynomial $$ p(x) = -11x^4+67x^3-72x^2+17x-1 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= \frac{ 1 }{ 11 }\\[1 em]x_3 &= 4.7913\\[1 em]x_4 &= 0.2087 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ -11x^4+67x^3-72x^2+17x-1 $.
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}{ 1 } $, with factors of 1.
The leading coefficient is $ \color{red}{ 11 }$, with factors of 1 and 11.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 1 }}{\text{ factors of 11 }} = \pm \dfrac{\text{ ( 1 ) }}{\text{ ( 1, 11 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} ~~ \pm \frac{ 1}{ 11} ~~ \end{aligned} $$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( 1 \right) = 0 $ so $ x = 1 $ 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-1 }$
$$ \frac{ -11x^4+67x^3-72x^2+17x-1}{ x-1} = -11x^3+56x^2-16x+1 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ -11x^4+67x^3-72x^2+17x-1}{ x-1} = -11x^3+56x^2-16x+1 $$Step 3:
The next rational root is $ x = \dfrac{ 1 }{ 11 } $
$$ \frac{ -11x^3+56x^2-16x+1}{ 11x-1} = -x^2+5x-1 $$Step 4:
The solutions of $ -x^2+5x-1 = 0 $ are: $ x = \dfrac{ 5 }{ 2 }-\dfrac{\sqrt{ 21 }}{ 2 } ~ \text{and} ~ x = \dfrac{ 5 }{ 2 }+\dfrac{\sqrt{ 21 }}{ 2 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.