Find roots of polynomial $$ p(x) = x^5-34x^3+29x^2+212x-300 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 2\\[1 em]x_2 &= -3\\[1 em]x_3 &= 2\\[1 em]x_4 &= 4.5249\\[1 em]x_5 &= -5.5249 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 2 } $ is a root of polynomial $ x^5-34x^3+29x^2+212x-300 $.
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}{ 300 } $, with a single factor of 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150 and 300.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 300 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 30}{ 1} \pm \frac{ 50}{ 1} \pm \frac{ 60}{ 1} \pm \frac{ 75}{ 1} \pm \frac{ 100}{ 1} \pm \frac{ 150}{ 1} \pm \frac{ 300}{ 1} ~~ \end{aligned} $$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( 2 \right) = 0 $ so $ x = 2 $ 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-2 }$
$$ \frac{ x^5-34x^3+29x^2+212x-300}{ x-2} = x^4+2x^3-30x^2-31x+150 $$Step 2:
The next rational root is $ x = 2 $
$$ \frac{ x^5-34x^3+29x^2+212x-300}{ x-2} = x^4+2x^3-30x^2-31x+150 $$Step 3:
The next rational root is $ x = -3 $
$$ \frac{ x^4+2x^3-30x^2-31x+150}{ x+3} = x^3-x^2-27x+50 $$Step 4:
The next rational root is $ x = 2 $
$$ \frac{ x^3-x^2-27x+50}{ x-2} = x^2+x-25 $$Step 5:
The solutions of $ x^2+x-25 = 0 $ are: $ x = -\dfrac{ 1 }{ 2 }-\dfrac{\sqrt{ 101 }}{ 2 } ~ \text{and} ~ x = -\dfrac{ 1 }{ 2 }+\dfrac{\sqrt{ 101 }}{ 2 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.