Find roots of polynomial $$ p(x) = \frac{5}{2}x^4+3x^3-\frac{13}{5}x^2-\frac{51}{10}x-\frac{28}{5} $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \frac{ 7 }{ 5 }\\[1 em]x_2 &= -\frac{ 8 }{ 5 }\\[1 em]x_3 &= -\frac{ 1 }{ 2 }+\frac{\sqrt{ 3 }}{ 2 }i\\[1 em]x_4 &= -\frac{ 1 }{ 2 }- \frac{\sqrt{ 3 }}{ 2 }i \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 10 } $.
$$ \begin{aligned} \frac{5}{2}x^4+3x^3-\frac{13}{5}x^2-\frac{51}{10}x-\frac{28}{5} & = 0 ~~~ / \cdot \color{blue}{ 10 } \\[1 em] 25x^4+30x^3-26x^2-51x-56 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 7 }{ 5 } } $ is a root of polynomial $ 25x^4+30x^3-26x^2-51x-56 $.
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}{ 56 } $, with factors of 1, 2, 4, 7, 8, 14, 28 and 56.
The leading coefficient is $ \color{red}{ 25 }$, with factors of 1, 5 and 25.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 56 }}{\text{ factors of 25 }} = \pm \dfrac{\text{ ( 1, 2, 4, 7, 8, 14, 28, 56 ) }}{\text{ ( 1, 5, 25 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 7}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 14}{ 1} \pm \frac{ 28}{ 1} \pm \frac{ 56}{ 1} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 2}{ 5} \pm \frac{ 4}{ 5} \pm \frac{ 7}{ 5} \pm \frac{ 8}{ 5} \pm \frac{ 14}{ 5} \pm \frac{ 28}{ 5} \pm \frac{ 56}{ 5} ~~ \pm \frac{ 1}{ 25} \pm \frac{ 2}{ 25} \pm \frac{ 4}{ 25} \pm \frac{ 7}{ 25} \pm \frac{ 8}{ 25} \pm \frac{ 14}{ 25} \pm \frac{ 28}{ 25} \pm \frac{ 56}{ 25} ~~ \end{aligned} $$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{ 7 }{ 5 } \right) = 0 $ so $ x = \dfrac{ 7 }{ 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-7 }$
$$ \frac{ 25x^4+30x^3-26x^2-51x-56}{ 5x-7} = 5x^3+13x^2+13x+8 $$Step 3:
The next rational root is $ x = \dfrac{ 7 }{ 5 } $
$$ \frac{ 25x^4+30x^3-26x^2-51x-56}{ 5x-7} = 5x^3+13x^2+13x+8 $$Step 4:
The next rational root is $ x = -\dfrac{ 8 }{ 5 } $
$$ \frac{ 5x^3+13x^2+13x+8}{ 5x+8} = x^2+x+1 $$Step 5:
The solutions of $ x^2+x+1 = 0 $ are: $ x = -\dfrac{ 1 }{ 2 }+\dfrac{\sqrt{ 3 }}{ 2 }i ~ \text{and} ~ x = -\dfrac{ 1 }{ 2 }-\dfrac{\sqrt{ 3 }}{ 2 }i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.