Find roots of polynomial $$ p(x) = -2x^5+19x^4-50x^3-21x^2+234x-216 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= 4\\[1 em]x_3 &= -2\\[1 em]x_4 &= \frac{ 3 }{ 2 }\\[1 em]x_5 &= 3 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ -2x^5+19x^4-50x^3-21x^2+234x-216 $.
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}{ 216 } $, with factors of 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108 and 216.
The leading coefficient is $ \color{red}{ 2 }$, with factors of 1 and 2.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 216 }}{\text{ factors of 2 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216 ) }}{\text{ ( 1, 2 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 27}{ 1} \pm \frac{ 36}{ 1} \pm \frac{ 54}{ 1} \pm \frac{ 72}{ 1} \pm \frac{ 108}{ 1} \pm \frac{ 216}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 8}{ 2} \pm \frac{ 9}{ 2} \pm \frac{ 12}{ 2} \pm \frac{ 18}{ 2} \pm \frac{ 24}{ 2} \pm \frac{ 27}{ 2} \pm \frac{ 36}{ 2} \pm \frac{ 54}{ 2} \pm \frac{ 72}{ 2} \pm \frac{ 108}{ 2} \pm \frac{ 216}{ 2} ~~ \end{aligned} $$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( 3 \right) = 0 $ so $ x = 3 $ 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-3 }$
$$ \frac{ -2x^5+19x^4-50x^3-21x^2+234x-216}{ x-3} = -2x^4+13x^3-11x^2-54x+72 $$Step 2:
The next rational root is $ x = 3 $
$$ \frac{ -2x^5+19x^4-50x^3-21x^2+234x-216}{ x-3} = -2x^4+13x^3-11x^2-54x+72 $$Step 3:
The next rational root is $ x = 4 $
$$ \frac{ -2x^4+13x^3-11x^2-54x+72}{ x-4} = -2x^3+5x^2+9x-18 $$Step 4:
The next rational root is $ x = -2 $
$$ \frac{ -2x^3+5x^2+9x-18}{ x+2} = -2x^2+9x-9 $$Step 5:
The next rational root is $ x = \dfrac{ 3 }{ 2 } $
$$ \frac{ -2x^2+9x-9}{ 2x-3} = -x+3 $$Step 6:
To find the last zero, solve equation $ -x+3 = 0 $
$$ \begin{aligned} -x+3 & = 0 \\[1 em] -1 \cdot x & = -3 \\[1 em] x & = - \frac{ 3 }{ -1 } \\[1 em] x & = 3 \end{aligned} $$