Find roots of polynomial $$ p(x) = 3x^5-14x^4+96x^3-360x^2+525x-250 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= 2\\[1 em]x_3 &= \frac{ 5 }{ 3 }\\[1 em]x_4 &= 5i\\[1 em]x_5 &= -5i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 3x^5-14x^4+96x^3-360x^2+525x-250 $.
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}{ 250 } $, with factors of 1, 2, 5, 10, 25, 50, 125 and 250.
The leading coefficient is $ \color{red}{ 3 }$, with factors of 1 and 3.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 250 }}{\text{ factors of 3 }} = \pm \dfrac{\text{ ( 1, 2, 5, 10, 25, 50, 125, 250 ) }}{\text{ ( 1, 3 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 50}{ 1} \pm \frac{ 125}{ 1} \pm \frac{ 250}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 5}{ 3} \pm \frac{ 10}{ 3} \pm \frac{ 25}{ 3} \pm \frac{ 50}{ 3} \pm \frac{ 125}{ 3} \pm \frac{ 250}{ 3} ~~ \end{aligned} $$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 }$
$$ \frac{ 3x^5-14x^4+96x^3-360x^2+525x-250}{ x-1} = 3x^4-11x^3+85x^2-275x+250 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ 3x^5-14x^4+96x^3-360x^2+525x-250}{ x-1} = 3x^4-11x^3+85x^2-275x+250 $$Step 3:
The next rational root is $ x = 2 $
$$ \frac{ 3x^4-11x^3+85x^2-275x+250}{ x-2} = 3x^3-5x^2+75x-125 $$Step 4:
The next rational root is $ x = \dfrac{ 5 }{ 3 } $
$$ \frac{ 3x^3-5x^2+75x-125}{ 3x-5} = x^2+25 $$Step 5:
The solutions of $ x^2+25 = 0 $ are: $ x = 5 i ~ \text{and} ~ x = -5 i $.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.