Find roots of polynomial $$ p(x) = 7x^4-50x^3+112x^2-78x+9 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= 3\\[1 em]x_3 &= \frac{ 1 }{ 7 }\\[1 em]x_4 &= 3 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 7x^4-50x^3+112x^2-78x+9 $.
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}{ 9 } $, with factors of 1, 3 and 9.
The leading coefficient is $ \color{red}{ 7 }$, with factors of 1 and 7.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 9 }}{\text{ factors of 7 }} = \pm \dfrac{\text{ ( 1, 3, 9 ) }}{\text{ ( 1, 7 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 9}{ 1} ~~ \pm \frac{ 1}{ 7} \pm \frac{ 3}{ 7} \pm \frac{ 9}{ 7} ~~ \end{aligned} $$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 }$
$$ \frac{ 7x^4-50x^3+112x^2-78x+9}{ x-1} = 7x^3-43x^2+69x-9 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ 7x^4-50x^3+112x^2-78x+9}{ x-1} = 7x^3-43x^2+69x-9 $$Step 3:
The next rational root is $ x = 3 $
$$ \frac{ 7x^3-43x^2+69x-9}{ x-3} = 7x^2-22x+3 $$Step 4:
The next rational root is $ x = \dfrac{ 1 }{ 7 } $
$$ \frac{ 7x^2-22x+3}{ 7x-1} = x-3 $$Step 5:
To find the last zero, solve equation $ x-3 = 0 $
$$ \begin{aligned} x-3 & = 0 \\[1 em] x & = 3 \end{aligned} $$