Find roots of polynomial $$ p(x) = 48x^3-380x^2+568x+765 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \frac{ 9 }{ 2 }\\[1 em]x_2 &= \frac{ 17 }{ 4 }\\[1 em]x_3 &= -\frac{ 5 }{ 6 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 9 }{ 2 } } $ is a root of polynomial $ 48x^3-380x^2+568x+765 $.
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}{ 765 } $, with factors of 1, 3, 5, 9, 15, 17, 45, 51, 85, 153, 255 and 765.
The leading coefficient is $ \color{red}{ 48 }$, with factors of 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 765 }}{\text{ factors of 48 }} = \pm \dfrac{\text{ ( 1, 3, 5, 9, 15, 17, 45, 51, 85, 153, 255, 765 ) }}{\text{ ( 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 17}{ 1} \pm \frac{ 45}{ 1} \pm \frac{ 51}{ 1} \pm \frac{ 85}{ 1} \pm \frac{ 153}{ 1} \pm \frac{ 255}{ 1} \pm \frac{ 765}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 9}{ 2} \pm \frac{ 15}{ 2} \pm \frac{ 17}{ 2} \pm \frac{ 45}{ 2} \pm \frac{ 51}{ 2} \pm \frac{ 85}{ 2} \pm \frac{ 153}{ 2} \pm \frac{ 255}{ 2} \pm \frac{ 765}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 5}{ 3} \pm \frac{ 9}{ 3} \pm \frac{ 15}{ 3} \pm \frac{ 17}{ 3} \pm \frac{ 45}{ 3} \pm \frac{ 51}{ 3} \pm \frac{ 85}{ 3} \pm \frac{ 153}{ 3} \pm \frac{ 255}{ 3} \pm \frac{ 765}{ 3} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 9}{ 4} \pm \frac{ 15}{ 4} \pm \frac{ 17}{ 4} \pm \frac{ 45}{ 4} \pm \frac{ 51}{ 4} \pm \frac{ 85}{ 4} \pm \frac{ 153}{ 4} \pm \frac{ 255}{ 4} \pm \frac{ 765}{ 4} ~~ \pm \frac{ 1}{ 6} \pm \frac{ 3}{ 6} \pm \frac{ 5}{ 6} \pm \frac{ 9}{ 6} \pm \frac{ 15}{ 6} \pm \frac{ 17}{ 6} \pm \frac{ 45}{ 6} \pm \frac{ 51}{ 6} \pm \frac{ 85}{ 6} \pm \frac{ 153}{ 6} \pm \frac{ 255}{ 6} \pm \frac{ 765}{ 6} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 3}{ 8} \pm \frac{ 5}{ 8} \pm \frac{ 9}{ 8} \pm \frac{ 15}{ 8} \pm \frac{ 17}{ 8} \pm \frac{ 45}{ 8} \pm \frac{ 51}{ 8} \pm \frac{ 85}{ 8} \pm \frac{ 153}{ 8} \pm \frac{ 255}{ 8} \pm \frac{ 765}{ 8} ~~ \pm \frac{ 1}{ 12} \pm \frac{ 3}{ 12} \pm \frac{ 5}{ 12} \pm \frac{ 9}{ 12} \pm \frac{ 15}{ 12} \pm \frac{ 17}{ 12} \pm \frac{ 45}{ 12} \pm \frac{ 51}{ 12} \pm \frac{ 85}{ 12} \pm \frac{ 153}{ 12} \pm \frac{ 255}{ 12} \pm \frac{ 765}{ 12} ~~ \pm \frac{ 1}{ 16} \pm \frac{ 3}{ 16} \pm \frac{ 5}{ 16} \pm \frac{ 9}{ 16} \pm \frac{ 15}{ 16} \pm \frac{ 17}{ 16} \pm \frac{ 45}{ 16} \pm \frac{ 51}{ 16} \pm \frac{ 85}{ 16} \pm \frac{ 153}{ 16} \pm \frac{ 255}{ 16} \pm \frac{ 765}{ 16} ~~ \pm \frac{ 1}{ 24} \pm \frac{ 3}{ 24} \pm \frac{ 5}{ 24} \pm \frac{ 9}{ 24} \pm \frac{ 15}{ 24} \pm \frac{ 17}{ 24} \pm \frac{ 45}{ 24} \pm \frac{ 51}{ 24} \pm \frac{ 85}{ 24} \pm \frac{ 153}{ 24} \pm \frac{ 255}{ 24} \pm \frac{ 765}{ 24} ~~ \pm \frac{ 1}{ 48} \pm \frac{ 3}{ 48} \pm \frac{ 5}{ 48} \pm \frac{ 9}{ 48} \pm \frac{ 15}{ 48} \pm \frac{ 17}{ 48} \pm \frac{ 45}{ 48} \pm \frac{ 51}{ 48} \pm \frac{ 85}{ 48} \pm \frac{ 153}{ 48} \pm \frac{ 255}{ 48} \pm \frac{ 765}{ 48} ~~ \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{ 9 }{ 2 } \right) = 0 $ so $ x = \dfrac{ 9 }{ 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-9 }$
$$ \frac{ 48x^3-380x^2+568x+765}{ 2x-9} = 24x^2-82x-85 $$Step 2:
The next rational root is $ x = \dfrac{ 9 }{ 2 } $
$$ \frac{ 48x^3-380x^2+568x+765}{ 2x-9} = 24x^2-82x-85 $$Step 3:
The next rational root is $ x = \dfrac{ 17 }{ 4 } $
$$ \frac{ 24x^2-82x-85}{ 4x-17} = 6x+5 $$Step 4:
To find the last zero, solve equation $ 6x+5 = 0 $
$$ \begin{aligned} 6x+5 & = 0 \\[1 em] 6 \cdot x & = -5 \\[1 em] x & = - \frac{ 5 }{ 6 } \end{aligned} $$