Find roots of polynomial $$ p(x) = 4t^4+26t^3-14t-4 $$
Answer
The roots of polynomial $ p(t) $ are:
$$ \begin{aligned}t_1 &= -\frac{ 1 }{ 2 }\\[1 em]t_2 &= -0.3868\\[1 em]t_3 &= 0.8056\\[1 em]t_4 &= -6.4188 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ t = -\dfrac{ 1 }{ 2 } } $ is a root of polynomial $ 4t^4+26t^3-14t-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}{ 4 }$, with factors of 1, 2 and 4.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 4 }}{\text{ factors of 4 }} = \pm \dfrac{\text{ ( 1, 2, 4 ) }}{\text{ ( 1, 2, 4 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 4}{ 4} ~~ \end{aligned} $$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 }{ 2 } \right) = 0 $ so $ x = -\dfrac{ 1 }{ 2 } $ 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}{ 2t+1 }$
$$ \frac{ 4t^4+26t^3-14t-4}{ 2t+1} = 2t^3+12t^2-6t-4 $$Step 2:
The next rational root is $ t = -\dfrac{ 1 }{ 2 } $
$$ \frac{ 4t^4+26t^3-14t-4}{ 2t+1} = 2t^3+12t^2-6t-4 $$Step 3:
Polynomial $ 2t^3+12t^2-6t-4 $ has no rational roots that can be found using Rational Root Test, so the roots were found using qubic formulas.