Find roots of polynomial $$ p(x) = 7x^5-49x^4+91x^3-63x^2+378 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= -1.2114\\[1 em]x_3 &= 4.3146\\[1 em]x_4 &= 0.4484+1.8008i\\[1 em]x_5 &= 0.4484-1.8008i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ 7x^5-49x^4+91x^3-63x^2+378 $.
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}{ 378 } $, with factors of 1, 2, 3, 6, 7, 9, 14, 18, 21, 27, 42, 54, 63, 126, 189 and 378.
The leading coefficient is $ \color{red}{ 7 }$, with factors of 1 and 7.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 378 }}{\text{ factors of 7 }} = \pm \dfrac{\text{ ( 1, 2, 3, 6, 7, 9, 14, 18, 21, 27, 42, 54, 63, 126, 189, 378 ) }}{\text{ ( 1, 7 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 7}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 14}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 21}{ 1} \pm \frac{ 27}{ 1} \pm \frac{ 42}{ 1} \pm \frac{ 54}{ 1} \pm \frac{ 63}{ 1} \pm \frac{ 126}{ 1} \pm \frac{ 189}{ 1} \pm \frac{ 378}{ 1} ~~ \pm \frac{ 1}{ 7} \pm \frac{ 2}{ 7} \pm \frac{ 3}{ 7} \pm \frac{ 6}{ 7} \pm \frac{ 7}{ 7} \pm \frac{ 9}{ 7} \pm \frac{ 14}{ 7} \pm \frac{ 18}{ 7} \pm \frac{ 21}{ 7} \pm \frac{ 27}{ 7} \pm \frac{ 42}{ 7} \pm \frac{ 54}{ 7} \pm \frac{ 63}{ 7} \pm \frac{ 126}{ 7} \pm \frac{ 189}{ 7} \pm \frac{ 378}{ 7} ~~ \end{aligned} $$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( 3 \right) = 0 $ so $ x = 3 $ 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-3 }$
$$ \frac{ 7x^5-49x^4+91x^3-63x^2+378}{ x-3} = 7x^4-28x^3+7x^2-42x-126 $$Step 2:
The next rational root is $ x = 3 $
$$ \frac{ 7x^5-49x^4+91x^3-63x^2+378}{ x-3} = 7x^4-28x^3+7x^2-42x-126 $$Step 3:
Polynomial $ 7x^4-28x^3+7x^2-42x-126 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.