Find roots of polynomial $$ p(x) = x^4-3x^3+8x^2-18x+12 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= 2\\[1 em]x_3 &= \sqrt{ 6 }i\\[1 em]x_4 &= -\sqrt{ 6 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ x^4-3x^3+8x^2-18x+12 $.
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}{ 12 } $, with a single factor of 1, 2, 3, 4, 6 and 12.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 12 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 12 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 12}{ 1} ~~ \end{aligned} $$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( 1 \right) = 0 $ so $ x = 1 $ 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-1 }$
$$ \frac{ x^4-3x^3+8x^2-18x+12}{ x-1} = x^3-2x^2+6x-12 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ x^4-3x^3+8x^2-18x+12}{ x-1} = x^3-2x^2+6x-12 $$Step 3:
The next rational root is $ x = 2 $
$$ \frac{ x^3-2x^2+6x-12}{ x-2} = x^2+6 $$Step 4:
The solutions of $ x^2+6 = 0 $ are: $ x = \sqrt{ 6 } i ~ \text{and} ~ x = -\sqrt{ 6 } i $.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.