Find roots of polynomial $$ p(x) = 3x^4-35x^3+\frac{99}{2}x^2+\frac{175}{2}x $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -1\\[1 em]x_3 &= 9.6416\\[1 em]x_4 &= 3.0251 \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 2 } $.
$$ \begin{aligned} 3x^4-35x^3+\frac{99}{2}x^2+\frac{175}{2}x & = 0 ~~~ / \cdot \color{blue}{ 2 } \\[1 em] 6x^4-70x^3+99x^2+175x & = 0 \end{aligned} $$Step 2:
Factor out $ \color{blue}{ x }$ from $ 6x^4-70x^3+99x^2+175x $ and solve two separate equations:
$$ \begin{aligned} 6x^4-70x^3+99x^2+175x & = 0\\[1 em] \color{blue}{ x }\cdot ( 6x^3-70x^2+99x+175 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 6x^3-70x^2+99x+175 & = 0 \end{aligned} $$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 $ 6x^3-70x^2+99x+175 $.
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}{ 175 } $, with factors of 1, 5, 7, 25, 35 and 175.
The leading coefficient is $ \color{red}{ 6 }$, with factors of 1, 2, 3 and 6.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 175 }}{\text{ factors of 6 }} = \pm \dfrac{\text{ ( 1, 5, 7, 25, 35, 175 ) }}{\text{ ( 1, 2, 3, 6 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 7}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 35}{ 1} \pm \frac{ 175}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 7}{ 2} \pm \frac{ 25}{ 2} \pm \frac{ 35}{ 2} \pm \frac{ 175}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 5}{ 3} \pm \frac{ 7}{ 3} \pm \frac{ 25}{ 3} \pm \frac{ 35}{ 3} \pm \frac{ 175}{ 3}\\[ 1 em] \pm \frac{ 1}{ 6} & \pm \frac{ 5}{ 6} & \pm \frac{ 7}{ 6} & \pm \frac{ 25}{ 6} & \pm \frac{ 35}{ 6} & \pm \frac{ 175}{ 6} ~~ \end{aligned} $$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 }$
$$ \frac{ 6x^3-70x^2+99x+175}{ x+1} = 6x^2-76x+175 $$Step 4:
The next rational root is $ x = -1 $
$$ \frac{ 6x^3-70x^2+99x+175}{ x+1} = 6x^2-76x+175 $$Step 5:
The solutions of $ 6x^2-76x+175 = 0 $ are: $ x = \dfrac{ 19 }{ 3 }-\dfrac{\sqrt{ 394 }}{ 6 } ~ \text{and} ~ x = \dfrac{ 19 }{ 3 }+\dfrac{\sqrt{ 394 }}{ 6 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.