Find roots of polynomial $$ p(x) = 2y^4+3y^3-16y^2+15y-4 $$
Answer
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= 1\\[1 em]y_2 &= -4\\[1 em]y_3 &= \frac{ 1 }{ 2 }\\[1 em]y_4 &= 1 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ y = 1 } $ is a root of polynomial $ 2y^4+3y^3-16y^2+15y-4 $.
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}{ 4 } $, with factors of 1, 2 and 4.
The leading coefficient is $ \color{red}{ 2 }$, with factors of 1 and 2.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 4 }}{\text{ factors of 2 }} = \pm \dfrac{\text{ ( 1, 2, 4 ) }}{\text{ ( 1, 2 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} ~~ \end{aligned} $$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}{ y-1 }$
$$ \frac{ 2y^4+3y^3-16y^2+15y-4}{ y-1} = 2y^3+5y^2-11y+4 $$Step 2:
The next rational root is $ y = 1 $
$$ \frac{ 2y^4+3y^3-16y^2+15y-4}{ y-1} = 2y^3+5y^2-11y+4 $$Step 3:
The next rational root is $ y = -4 $
$$ \frac{ 2y^3+5y^2-11y+4}{ y+4} = 2y^2-3y+1 $$Step 4:
The next rational root is $ y = \dfrac{ 1 }{ 2 } $
$$ \frac{ 2y^2-3y+1}{ 2y-1} = y-1 $$Step 5:
To find the last zero, solve equation $ y-1 = 0 $
$$ \begin{aligned} y-1 & = 0 \\[1 em] y & = 1 \end{aligned} $$