Find roots of polynomial $$ p(x) = 8x^4-14x^3-127x^2-24x+45 $$

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

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

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 5 } $ is a root of polynomial $ 8x^4-14x^3-127x^2-24x+45 $.

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}{ 45 } $, with factors of 1, 3, 5, 9, 15 and 45.

The leading coefficient is $ \color{red}{ 8 }$, with factors of 1, 2, 4 and 8.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 45 }}{\text{ factors of 8 }} = \pm \dfrac{\text{ ( 1, 3, 5, 9, 15, 45 ) }}{\text{ ( 1, 2, 4, 8 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 45}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 9}{ 2} \pm \frac{ 15}{ 2} \pm \frac{ 45}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 9}{ 4} \pm \frac{ 15}{ 4} \pm \frac{ 45}{ 4}\\[ 1 em] \pm \frac{ 1}{ 8} & \pm \frac{ 3}{ 8} & \pm \frac{ 5}{ 8} & \pm \frac{ 9}{ 8} & \pm \frac{ 15}{ 8} & \pm \frac{ 45}{ 8} ~~ \end{aligned} $$

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( 5 \right) = 0 $ so $ x = 5 $ 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-5 }$

$$ \frac{ 8x^4-14x^3-127x^2-24x+45}{ x-5} = 8x^3+26x^2+3x-9 $$

Step 2:

The next rational root is $ x = 5 $

$$ \frac{ 8x^4-14x^3-127x^2-24x+45}{ x-5} = 8x^3+26x^2+3x-9 $$

Step 3:

The next rational root is $ x = -3 $

$$ \frac{ 8x^3+26x^2+3x-9}{ x+3} = 8x^2+2x-3 $$

Step 4:

The next rational root is $ x = \dfrac{ 1 }{ 2 } $

$$ \frac{ 8x^2+2x-3}{ 2x-1} = 4x+3 $$

Step 5:

To find the last zero, solve equation $ 4x+3 = 0 $

$$ \begin{aligned} 4x+3 & = 0 \\[1 em] 4 \cdot x & = -3 \\[1 em] x & = - \frac{ 3 }{ 4 } \end{aligned} $$

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