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