Find roots of polynomial $$ p(x) = 6x^4-21x^3-4x^2+24x-35 $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= \frac{ 7 }{ 2 }\\[1 em]x_2 &= -1.3717\\[1 em]x_3 &= 0.6859+0.8629i\\[1 em]x_4 &= 0.6859-0.8629i \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 7 }{ 2 } } $ is a root of polynomial $ 6x^4-21x^3-4x^2+24x-35 $.

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}{ 35 } $, with factors of 1, 5, 7 and 35.

The leading coefficient is $ \color{red}{ 6 }$, with factors of 1, 2, 3 and 6.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 35 }}{\text{ factors of 6 }} = \pm \dfrac{\text{ ( 1, 5, 7, 35 ) }}{\text{ ( 1, 2, 3, 6 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 7}{ 1} \pm \frac{ 35}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 7}{ 2} \pm \frac{ 35}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 5}{ 3} \pm \frac{ 7}{ 3} \pm \frac{ 35}{ 3} ~~ \pm \frac{ 1}{ 6} \pm \frac{ 5}{ 6} \pm \frac{ 7}{ 6} \pm \frac{ 35}{ 6}\\[ 1 em] \end{aligned} $$

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( \dfrac{ 7 }{ 2 } \right) = 0 $ so $ x = \dfrac{ 7 }{ 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-7 }$

$$ \frac{ 6x^4-21x^3-4x^2+24x-35}{ 2x-7} = 3x^3-2x+5 $$

Step 2:

The next rational root is $ x = \dfrac{ 7 }{ 2 } $

$$ \frac{ 6x^4-21x^3-4x^2+24x-35}{ 2x-7} = 3x^3-2x+5 $$

Step 3:

Polynomial $ 3x^3-2x+5 $ has no rational roots that can be found using Rational Root Test, so the roots were found using qubic formulas.

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