Find roots of polynomial $$ p(x) = 2x^5-9x^4+15x^3-11x^2+3x $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 1\\[1 em]x_3 &= \frac{ 3 }{ 2 }\\[1 em]x_4 &= 1\\[1 em]x_5 &= 1 \end{aligned} $$Explanation
Step 1:
Factor out $ \color{blue}{ x }$ from $ 2x^5-9x^4+15x^3-11x^2+3x $ and solve two separate equations:
$$ \begin{aligned} 2x^5-9x^4+15x^3-11x^2+3x & = 0\\[1 em] \color{blue}{ x }\cdot ( 2x^4-9x^3+15x^2-11x+3 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 2x^4-9x^3+15x^2-11x+3 & = 0 \end{aligned} $$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 = 1 } $ is a root of polynomial $ 2x^4-9x^3+15x^2-11x+3 $.
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}{ 3 } $, with factors of 1 and 3.
The leading coefficient is $ \color{red}{ 2 }$, with factors of 1 and 2.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 3 }}{\text{ factors of 2 }} = \pm \dfrac{\text{ ( 1, 3 ) }}{\text{ ( 1, 2 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} ~~ \end{aligned} $$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( 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 }$
$$ \frac{ 2x^4-9x^3+15x^2-11x+3}{ x-1} = 2x^3-7x^2+8x-3 $$Step 3:
The next rational root is $ x = 1 $
$$ \frac{ 2x^4-9x^3+15x^2-11x+3}{ x-1} = 2x^3-7x^2+8x-3 $$Step 4:
The next rational root is $ x = \dfrac{ 3 }{ 2 } $
$$ \frac{ 2x^3-7x^2+8x-3}{ 2x-3} = x^2-2x+1 $$Step 5:
The next rational root is $ x = 1 $
$$ \frac{ x^2-2x+1}{ x-1} = x-1 $$Step 6:
To find the last zero, solve equation $ x-1 = 0 $
$$ \begin{aligned} x-1 & = 0 \\[1 em] x & = 1 \end{aligned} $$