Find roots of polynomial $$ p(x) = x^4+x^3-32x^2-12x+128 $$

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

$$ \begin{aligned}x_1 &= 2\\[1 em]x_2 &= -2.3208\\[1 em]x_3 &= 4.9227\\[1 em]x_4 &= -5.6019 \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 2 } $ is a root of polynomial $ x^4+x^3-32x^2-12x+128 $.

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}{ 128 } $, with a single factor of 1, 2, 4, 8, 16, 32, 64 and 128.

The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 128 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 4, 8, 16, 32, 64, 128 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 32}{ 1} \pm \frac{ 64}{ 1} \pm \frac{ 128}{ 1} ~~ \end{aligned} $$

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 }$

$$ \frac{ x^4+x^3-32x^2-12x+128}{ x-2} = x^3+3x^2-26x-64 $$

Step 2:

The next rational root is $ x = 2 $

$$ \frac{ x^4+x^3-32x^2-12x+128}{ x-2} = x^3+3x^2-26x-64 $$

Step 3:

Polynomial $ x^3+3x^2-26x-64 $ has no rational roots that can be found using Rational Root Test, so the roots were found using qubic formulas.

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