Find roots of polynomial $$ p(x) = 20x^3+8x^2-83x+55 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= -\frac{ 5 }{ 2 }\\[1 em]x_3 &= \frac{ 11 }{ 10 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 20x^3+8x^2-83x+55 $.
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}{ 55 } $, with factors of 1, 5, 11 and 55.
The leading coefficient is $ \color{red}{ 20 }$, with factors of 1, 2, 4, 5, 10 and 20.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 55 }}{\text{ factors of 20 }} = \pm \dfrac{\text{ ( 1, 5, 11, 55 ) }}{\text{ ( 1, 2, 4, 5, 10, 20 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 11}{ 1} \pm \frac{ 55}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 11}{ 2} \pm \frac{ 55}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 11}{ 4} \pm \frac{ 55}{ 4} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 5}{ 5} \pm \frac{ 11}{ 5} \pm \frac{ 55}{ 5} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 5}{ 10} \pm \frac{ 11}{ 10} \pm \frac{ 55}{ 10} ~~ \pm \frac{ 1}{ 20} \pm \frac{ 5}{ 20} \pm \frac{ 11}{ 20} \pm \frac{ 55}{ 20} ~~ \end{aligned} $$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 }$
$$ \frac{ 20x^3+8x^2-83x+55}{ x-1} = 20x^2+28x-55 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ 20x^3+8x^2-83x+55}{ x-1} = 20x^2+28x-55 $$Step 3:
The next rational root is $ x = -\dfrac{ 5 }{ 2 } $
$$ \frac{ 20x^2+28x-55}{ 2x+5} = 10x-11 $$Step 4:
To find the last zero, solve equation $ 10x-11 = 0 $
$$ \begin{aligned} 10x-11 & = 0 \\[1 em] 10 \cdot x & = 11 \\[1 em] x & = \frac{ 11 }{ 10 } \end{aligned} $$