Find roots of polynomial $$ p(x) = x^5-2x^4-28x^3+50x^2+75x $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 3\\[1 em]x_3 &= 5\\[1 em]x_4 &= -1\\[1 em]x_5 &= -5 \end{aligned} $$Explanation
Step 1:
Factor out $ \color{blue}{ x }$ from $ x^5-2x^4-28x^3+50x^2+75x $ and solve two separate equations:
$$ \begin{aligned} x^5-2x^4-28x^3+50x^2+75x & = 0\\[1 em] \color{blue}{ x }\cdot ( x^4-2x^3-28x^2+50x+75 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x^4-2x^3-28x^2+50x+75 & = 0 \end{aligned} $$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 = 3 } $ is a root of polynomial $ x^4-2x^3-28x^2+50x+75 $.
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}{ 75 } $, with a single factor of 1, 3, 5, 15, 25 and 75.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 75 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 3, 5, 15, 25, 75 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 75}{ 1} ~~ \end{aligned} $$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( 3 \right) = 0 $ so $ x = 3 $ 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-3 }$
$$ \frac{ x^4-2x^3-28x^2+50x+75}{ x-3} = x^3+x^2-25x-25 $$Step 3:
The next rational root is $ x = 3 $
$$ \frac{ x^4-2x^3-28x^2+50x+75}{ x-3} = x^3+x^2-25x-25 $$Step 4:
The next rational root is $ x = 5 $
$$ \frac{ x^3+x^2-25x-25}{ x-5} = x^2+6x+5 $$Step 5:
The next rational root is $ x = -1 $
$$ \frac{ x^2+6x+5}{ x+1} = x+5 $$Step 6:
To find the last zero, solve equation $ x+5 = 0 $
$$ \begin{aligned} x+5 & = 0 \\[1 em] x & = -5 \end{aligned} $$