Find roots of polynomial $$ p(x) = x^4-2x^3-14x^2+40x $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -4\\[1 em]x_3 &= 3+i\\[1 em]x_4 &= 3-i \end{aligned} $$Explanation
Step 1:
Factor out $ \color{blue}{ x }$ from $ x^4-2x^3-14x^2+40x $ and solve two separate equations:
$$ \begin{aligned} x^4-2x^3-14x^2+40x & = 0\\[1 em] \color{blue}{ x }\cdot ( x^3-2x^2-14x+40 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x^3-2x^2-14x+40 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
Use rational root test to find out that the $ \color{blue}{ x = -4 } $ is a root of polynomial $ x^3-2x^2-14x+40 $.
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}{ 40 } $, with a single factor of 1, 2, 4, 5, 8, 10, 20 and 40.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 40 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 4, 5, 8, 10, 20, 40 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 40}{ 1} ~~ \end{aligned} $$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( -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 }$
$$ \frac{ x^3-2x^2-14x+40}{ x+4} = x^2-6x+10 $$Step 3:
The next rational root is $ x = -4 $
$$ \frac{ x^3-2x^2-14x+40}{ x+4} = x^2-6x+10 $$Step 4:
The solutions of $ x^2-6x+10 = 0 $ are: $ x = 3+i ~ \text{and} ~ x = 3-i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.