Find roots of polynomial $$ p(x) = 2x^4-5x^3-17x^2+32x-48 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 4\\[1 em]x_2 &= -3\\[1 em]x_3 &= \frac{ 3 }{ 4 }+\frac{\sqrt{ 23 }}{ 4 }i\\[1 em]x_4 &= \frac{ 3 }{ 4 }- \frac{\sqrt{ 23 }}{ 4 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 4 } $ is a root of polynomial $ 2x^4-5x^3-17x^2+32x-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( 4 \right) = 0 $ so $ x = 4 $ 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-4 }$
$$ \frac{ 2x^4-5x^3-17x^2+32x-48}{ x-4} = 2x^3+3x^2-5x+12 $$Step 2:
The next rational root is $ x = 4 $
$$ \frac{ 2x^4-5x^3-17x^2+32x-48}{ x-4} = 2x^3+3x^2-5x+12 $$Step 3:
The next rational root is $ x = -3 $
$$ \frac{ 2x^3+3x^2-5x+12}{ x+3} = 2x^2-3x+4 $$Step 4:
The solutions of $ 2x^2-3x+4 = 0 $ are: $ x = \dfrac{ 3 }{ 4 }+\dfrac{\sqrt{ 23 }}{ 4 }i ~ \text{and} ~ x = \dfrac{ 3 }{ 4 }-\dfrac{\sqrt{ 23 }}{ 4 }i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.