Find roots of polynomial $$ p(x) = y^5+5y^4+10y^3+10y^2+5y+1 $$
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= -1\\[1 em]y_2 &= -1\\[1 em]y_3 &= -1\\[1 em]y_4 &= -1\\[1 em]y_5 &= -1 \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ y = -1 } $ is a root of polynomial $ y^5+5y^4+10y^3+10y^2+5y+1 $.
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}{ 1 } $, with a single factor of 1.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 1 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} ~~ \end{aligned} $$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}{ y+1 }$
$$ \frac{ y^5+5y^4+10y^3+10y^2+5y+1}{ y+1} = y^4+4y^3+6y^2+4y+1 $$Step 2:
The next rational root is $ y = -1 $
$$ \frac{ y^5+5y^4+10y^3+10y^2+5y+1}{ y+1} = y^4+4y^3+6y^2+4y+1 $$Step 3:
The next rational root is $ y = -1 $
$$ \frac{ y^4+4y^3+6y^2+4y+1}{ y+1} = y^3+3y^2+3y+1 $$Step 4:
The next rational root is $ y = -1 $
$$ \frac{ y^3+3y^2+3y+1}{ y+1} = y^2+2y+1 $$Step 5:
The next rational root is $ y = -1 $
$$ \frac{ y^2+2y+1}{ y+1} = y+1 $$Step 6:
To find the last zero, solve equation $ y+1 = 0 $
$$ \begin{aligned} y+1 & = 0 \\[1 em] y & = -1 \end{aligned} $$