Find roots of polynomial $$ p(x) = 64y^4+44y^2-3 $$
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= \dfrac{ 1 }{ 4 }\\[1 em]y_2 &= -\dfrac{ 1 }{ 4 }\\[1 em]y_3 &= \dfrac{\sqrt{ 3 }}{ 2 }i\\[1 em]y_4 &= - \dfrac{\sqrt{ 3 }}{ 2 }i \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ y = \dfrac{ 1 }{ 4 } } $ is a root of polynomial $ 64y^4+44y^2-3 $.
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}{ 64 }$, with factors of 1, 2, 4, 8, 16, 32 and 64.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 3 }}{\text{ factors of 64 }} = \pm \dfrac{\text{ ( 1, 3 ) }}{\text{ ( 1, 2, 4, 8, 16, 32, 64 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 3}{ 8} ~~ \pm \frac{ 1}{ 16} \pm \frac{ 3}{ 16} ~~ \pm \frac{ 1}{ 32} \pm \frac{ 3}{ 32} ~~ \pm \frac{ 1}{ 64} \pm \frac{ 3}{ 64} ~~ \end{aligned} $$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( \dfrac{ 1 }{ 4 } \right) = 0 $ so $ x = \dfrac{ 1 }{ 4 } $ 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}{ 4y-1 }$
$$ \frac{ 64y^4+44y^2-3}{ 4y-1} = 16y^3+4y^2+12y+3 $$Step 2:
The next rational root is $ y = \dfrac{ 1 }{ 4 } $
$$ \frac{ 64y^4+44y^2-3}{ 4y-1} = 16y^3+4y^2+12y+3 $$Step 3:
The next rational root is $ y = -\dfrac{ 1 }{ 4 } $
$$ \frac{ 16y^3+4y^2+12y+3}{ 4y+1} = 4y^2+3 $$Step 4:
The solutions of $ 4y^2+3 = 0 $ are: $ y = \dfrac{\sqrt{ 3 }}{ 2 } i ~ \text{and} ~ y = - \dfrac{\sqrt{ 3 }}{ 2 } i $.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.