Find roots of polynomial $$ p(x) = x^4-\frac{11}{10}x^3+\frac{6}{25}x^2-\frac{7}{50}x $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 1\\[1 em]x_3 &= \frac{ 1 }{ 20 }+\frac{\sqrt{ 55 }}{ 20 }i\\[1 em]x_4 &= \frac{ 1 }{ 20 }- \frac{\sqrt{ 55 }}{ 20 }i \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 50 } $.
$$ \begin{aligned} x^4-\frac{11}{10}x^3+\frac{6}{25}x^2-\frac{7}{50}x & = 0 ~~~ / \cdot \color{blue}{ 50 } \\[1 em] 50x^4-55x^3+12x^2-7x & = 0 \end{aligned} $$Step 2:
Factor out $ \color{blue}{ x }$ from $ 50x^4-55x^3+12x^2-7x $ and solve two separate equations:
$$ \begin{aligned} 50x^4-55x^3+12x^2-7x & = 0\\[1 em] \color{blue}{ x }\cdot ( 50x^3-55x^2+12x-7 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 50x^3-55x^2+12x-7 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 3:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 50x^3-55x^2+12x-7 $.
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}{ 7 } $, with factors of 1 and 7.
The leading coefficient is $ \color{red}{ 50 }$, with factors of 1, 2, 5, 10, 25 and 50.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 7 }}{\text{ factors of 50 }} = \pm \dfrac{\text{ ( 1, 7 ) }}{\text{ ( 1, 2, 5, 10, 25, 50 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 7}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 7}{ 2} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 7}{ 5} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 7}{ 10} ~~ \pm \frac{ 1}{ 25} \pm \frac{ 7}{ 25} ~~ \pm \frac{ 1}{ 50} \pm \frac{ 7}{ 50} ~~ \end{aligned} $$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 }$
$$ \frac{ 50x^3-55x^2+12x-7}{ x-1} = 50x^2-5x+7 $$Step 4:
The next rational root is $ x = 1 $
$$ \frac{ 50x^3-55x^2+12x-7}{ x-1} = 50x^2-5x+7 $$Step 5:
The solutions of $ 50x^2-5x+7 = 0 $ are: $ x = \dfrac{ 1 }{ 20 }+\dfrac{\sqrt{ 55 }}{ 20 }i ~ \text{and} ~ x = \dfrac{ 1 }{ 20 }-\dfrac{\sqrt{ 55 }}{ 20 }i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.