Find roots of polynomial $$ p(x) = 3x^5+2x^4-87x^3-58x^2+300x+200 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 2\\[1 em]x_2 &= 5\\[1 em]x_3 &= -2\\[1 em]x_4 &= -5\\[1 em]x_5 &= -\frac{ 2 }{ 3 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 2 } $ is a root of polynomial $ 3x^5+2x^4-87x^3-58x^2+300x+200 $.
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}{ 200 } $, with factors of 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100 and 200.
The leading coefficient is $ \color{red}{ 3 }$, with factors of 1 and 3.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 200 }}{\text{ factors of 3 }} = \pm \dfrac{\text{ ( 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200 ) }}{\text{ ( 1, 3 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 40}{ 1} \pm \frac{ 50}{ 1} \pm \frac{ 100}{ 1} \pm \frac{ 200}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 5}{ 3} \pm \frac{ 8}{ 3} \pm \frac{ 10}{ 3} \pm \frac{ 20}{ 3} \pm \frac{ 25}{ 3} \pm \frac{ 40}{ 3} \pm \frac{ 50}{ 3} \pm \frac{ 100}{ 3} \pm \frac{ 200}{ 3} ~~ \end{aligned} $$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 }$
$$ \frac{ 3x^5+2x^4-87x^3-58x^2+300x+200}{ x-2} = 3x^4+8x^3-71x^2-200x-100 $$Step 2:
The next rational root is $ x = 2 $
$$ \frac{ 3x^5+2x^4-87x^3-58x^2+300x+200}{ x-2} = 3x^4+8x^3-71x^2-200x-100 $$Step 3:
The next rational root is $ x = 5 $
$$ \frac{ 3x^4+8x^3-71x^2-200x-100}{ x-5} = 3x^3+23x^2+44x+20 $$Step 4:
The next rational root is $ x = -2 $
$$ \frac{ 3x^3+23x^2+44x+20}{ x+2} = 3x^2+17x+10 $$Step 5:
The next rational root is $ x = -5 $
$$ \frac{ 3x^2+17x+10}{ x+5} = 3x+2 $$Step 6:
To find the last zero, solve equation $ 3x+2 = 0 $
$$ \begin{aligned} 3x+2 & = 0 \\[1 em] 3 \cdot x & = -2 \\[1 em] x & = - \frac{ 2 }{ 3 } \end{aligned} $$