Find roots of polynomial $$ p(x) = 2x^4-5x^3-35x^2+80x+48 $$

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

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

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ 2x^4-5x^3-35x^2+80x+48 $.

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}{ 48 } $, with factors of 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.

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

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 48 }}{\text{ factors of 2 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 ) }}{\text{ ( 1, 2 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 48}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 8}{ 2} \pm \frac{ 12}{ 2} \pm \frac{ 16}{ 2} \pm \frac{ 24}{ 2} \pm \frac{ 48}{ 2} ~~ \end{aligned} $$

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 }$

$$ \frac{ 2x^4-5x^3-35x^2+80x+48}{ x-3} = 2x^3+x^2-32x-16 $$

Step 2:

The next rational root is $ x = 3 $

$$ \frac{ 2x^4-5x^3-35x^2+80x+48}{ x-3} = 2x^3+x^2-32x-16 $$

Step 3:

The next rational root is $ x = 4 $

$$ \frac{ 2x^3+x^2-32x-16}{ x-4} = 2x^2+9x+4 $$

Step 4:

The next rational root is $ x = -4 $

$$ \frac{ 2x^2+9x+4}{ x+4} = 2x+1 $$

Step 5:

To find the last zero, solve equation $ 2x+1 = 0 $

$$ \begin{aligned} 2x+1 & = 0 \\[1 em] 2 \cdot x & = -1 \\[1 em] x & = - \frac{ 1 }{ 2 } \end{aligned} $$

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