Find roots of polynomial $$ p(x) = 180x^6+1317x^5+2978x^4+2617x^3+1080x^2+212x+16 $$

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

$$ \begin{aligned}x_1 &= -2\\[1 em]x_2 &= -4\\[1 em]x_3 &= -\frac{ 1 }{ 3 }\\[1 em]x_4 &= -\frac{ 1 }{ 4 }\\[1 em]x_5 &= -\frac{ 2 }{ 5 }\\[1 em]x_6 &= -\frac{ 1 }{ 3 } \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = -2 } $ is a root of polynomial $ 180x^6+1317x^5+2978x^4+2617x^3+1080x^2+212x+16 $.

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}{ 16 } $, with factors of 1, 2, 4, 8 and 16.

The leading coefficient is $ \color{red}{ 180 }$, with factors of 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90 and 180.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 16 }}{\text{ factors of 180 }} = \pm \dfrac{\text{ ( 1, 2, 4, 8, 16 ) }}{\text{ ( 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 16}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 8}{ 2} \pm \frac{ 16}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 8}{ 3} \pm \frac{ 16}{ 3} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 4}{ 4} \pm \frac{ 8}{ 4} \pm \frac{ 16}{ 4} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 2}{ 5} \pm \frac{ 4}{ 5} \pm \frac{ 8}{ 5} \pm \frac{ 16}{ 5} ~~ \pm \frac{ 1}{ 6} \pm \frac{ 2}{ 6} \pm \frac{ 4}{ 6} \pm \frac{ 8}{ 6} \pm \frac{ 16}{ 6} ~~ \pm \frac{ 1}{ 9} \pm \frac{ 2}{ 9} \pm \frac{ 4}{ 9} \pm \frac{ 8}{ 9} \pm \frac{ 16}{ 9} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 2}{ 10} \pm \frac{ 4}{ 10} \pm \frac{ 8}{ 10} \pm \frac{ 16}{ 10} ~~ \pm \frac{ 1}{ 12} \pm \frac{ 2}{ 12} \pm \frac{ 4}{ 12} \pm \frac{ 8}{ 12} \pm \frac{ 16}{ 12} ~~ \pm \frac{ 1}{ 15} \pm \frac{ 2}{ 15} \pm \frac{ 4}{ 15} \pm \frac{ 8}{ 15} \pm \frac{ 16}{ 15} ~~ \pm \frac{ 1}{ 18} \pm \frac{ 2}{ 18} \pm \frac{ 4}{ 18} \pm \frac{ 8}{ 18} \pm \frac{ 16}{ 18} ~~ \pm \frac{ 1}{ 20} \pm \frac{ 2}{ 20} \pm \frac{ 4}{ 20} \pm \frac{ 8}{ 20} \pm \frac{ 16}{ 20} ~~ \pm \frac{ 1}{ 30} \pm \frac{ 2}{ 30} \pm \frac{ 4}{ 30} \pm \frac{ 8}{ 30} \pm \frac{ 16}{ 30} ~~ \pm \frac{ 1}{ 36} \pm \frac{ 2}{ 36} \pm \frac{ 4}{ 36} \pm \frac{ 8}{ 36} \pm \frac{ 16}{ 36} ~~ \pm \frac{ 1}{ 45} \pm \frac{ 2}{ 45} \pm \frac{ 4}{ 45} \pm \frac{ 8}{ 45} \pm \frac{ 16}{ 45} ~~ \pm \frac{ 1}{ 60} \pm \frac{ 2}{ 60} \pm \frac{ 4}{ 60} \pm \frac{ 8}{ 60} \pm \frac{ 16}{ 60} ~~ \pm \frac{ 1}{ 90} \pm \frac{ 2}{ 90} \pm \frac{ 4}{ 90} \pm \frac{ 8}{ 90} \pm \frac{ 16}{ 90} ~~ \pm \frac{ 1}{ 180} \pm \frac{ 2}{ 180} \pm \frac{ 4}{ 180} \pm \frac{ 8}{ 180} \pm \frac{ 16}{ 180} ~~ \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( -2 \right) = 0 $ so $ x = -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}{ x+2 }$

$$ \frac{ 180x^6+1317x^5+2978x^4+2617x^3+1080x^2+212x+16}{ x+2} = 180x^5+957x^4+1064x^3+489x^2+102x+8 $$

Step 2:

The next rational root is $ x = -2 $

$$ \frac{ 180x^6+1317x^5+2978x^4+2617x^3+1080x^2+212x+16}{ x+2} = 180x^5+957x^4+1064x^3+489x^2+102x+8 $$

Step 3:

The next rational root is $ x = -4 $

$$ \frac{ 180x^5+957x^4+1064x^3+489x^2+102x+8}{ x+4} = 180x^4+237x^3+116x^2+25x+2 $$

Step 4:

The next rational root is $ x = -\dfrac{ 1 }{ 3 } $

$$ \frac{ 180x^4+237x^3+116x^2+25x+2}{ 3x+1} = 60x^3+59x^2+19x+2 $$

Step 5:

The next rational root is $ x = -\dfrac{ 1 }{ 4 } $

$$ \frac{ 60x^3+59x^2+19x+2}{ 4x+1} = 15x^2+11x+2 $$

Step 6:

The next rational root is $ x = -\dfrac{ 2 }{ 5 } $

$$ \frac{ 15x^2+11x+2}{ 5x+2} = 3x+1 $$

Step 7:

To find the last zero, solve equation $ 3x+1 = 0 $

$$ \begin{aligned} 3x+1 & = 0 \\[1 em] 3 \cdot x & = -1 \\[1 em] x & = - \frac{ 1 }{ 3 } \end{aligned} $$

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