Find roots of polynomial $$ p(x) = 10x^4-51x^3+39x^2+107x-105 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= 3\\[1 em]x_3 &= \frac{ 5 }{ 2 }\\[1 em]x_4 &= -\frac{ 7 }{ 5 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 10x^4-51x^3+39x^2+107x-105 $.
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}{ 105 } $, with factors of 1, 3, 5, 7, 15, 21, 35 and 105.
The leading coefficient is $ \color{red}{ 10 }$, with factors of 1, 2, 5 and 10.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 105 }}{\text{ factors of 10 }} = \pm \dfrac{\text{ ( 1, 3, 5, 7, 15, 21, 35, 105 ) }}{\text{ ( 1, 2, 5, 10 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 7}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 21}{ 1} \pm \frac{ 35}{ 1} \pm \frac{ 105}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 7}{ 2} \pm \frac{ 15}{ 2} \pm \frac{ 21}{ 2} \pm \frac{ 35}{ 2} \pm \frac{ 105}{ 2} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 3}{ 5} \pm \frac{ 5}{ 5} \pm \frac{ 7}{ 5} \pm \frac{ 15}{ 5} \pm \frac{ 21}{ 5} \pm \frac{ 35}{ 5} \pm \frac{ 105}{ 5} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 3}{ 10} \pm \frac{ 5}{ 10} \pm \frac{ 7}{ 10} \pm \frac{ 15}{ 10} \pm \frac{ 21}{ 10} \pm \frac{ 35}{ 10} \pm \frac{ 105}{ 10}\\[ 1 em] \end{aligned} $$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 }$
$$ \frac{ 10x^4-51x^3+39x^2+107x-105}{ x-1} = 10x^3-41x^2-2x+105 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ 10x^4-51x^3+39x^2+107x-105}{ x-1} = 10x^3-41x^2-2x+105 $$Step 3:
The next rational root is $ x = 3 $
$$ \frac{ 10x^3-41x^2-2x+105}{ x-3} = 10x^2-11x-35 $$Step 4:
The next rational root is $ x = \dfrac{ 5 }{ 2 } $
$$ \frac{ 10x^2-11x-35}{ 2x-5} = 5x+7 $$Step 5:
To find the last zero, solve equation $ 5x+7 = 0 $
$$ \begin{aligned} 5x+7 & = 0 \\[1 em] 5 \cdot x & = -7 \\[1 em] x & = - \frac{ 7 }{ 5 } \end{aligned} $$