Find roots of polynomial $$ p(x) = 16x^4-1305x^2+729 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 9\\[1 em]x_2 &= -9\\[1 em]x_3 &= \frac{ 3 }{ 4 }\\[1 em]x_4 &= -\frac{ 3 }{ 4 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 9 } $ is a root of polynomial $ 16x^4-1305x^2+729 $.
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}{ 729 } $, with factors of 1, 3, 9, 27, 81, 243 and 729.
The leading coefficient is $ \color{red}{ 16 }$, with factors of 1, 2, 4, 8 and 16.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 729 }}{\text{ factors of 16 }} = \pm \dfrac{\text{ ( 1, 3, 9, 27, 81, 243, 729 ) }}{\text{ ( 1, 2, 4, 8, 16 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 27}{ 1} \pm \frac{ 81}{ 1} \pm \frac{ 243}{ 1} \pm \frac{ 729}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 9}{ 2} \pm \frac{ 27}{ 2} \pm \frac{ 81}{ 2} \pm \frac{ 243}{ 2} \pm \frac{ 729}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 9}{ 4} \pm \frac{ 27}{ 4} \pm \frac{ 81}{ 4} \pm \frac{ 243}{ 4} \pm \frac{ 729}{ 4} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 3}{ 8} \pm \frac{ 9}{ 8} \pm \frac{ 27}{ 8} \pm \frac{ 81}{ 8} \pm \frac{ 243}{ 8} \pm \frac{ 729}{ 8} ~~ \pm \frac{ 1}{ 16} \pm \frac{ 3}{ 16} \pm \frac{ 9}{ 16} \pm \frac{ 27}{ 16} \pm \frac{ 81}{ 16} \pm \frac{ 243}{ 16} \pm \frac{ 729}{ 16} ~~ \end{aligned} $$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( 9 \right) = 0 $ so $ x = 9 $ 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-9 }$
$$ \frac{ 16x^4-1305x^2+729}{ x-9} = 16x^3+144x^2-9x-81 $$Step 2:
The next rational root is $ x = 9 $
$$ \frac{ 16x^4-1305x^2+729}{ x-9} = 16x^3+144x^2-9x-81 $$Step 3:
The next rational root is $ x = -9 $
$$ \frac{ 16x^3+144x^2-9x-81}{ x+9} = 16x^2-9 $$Step 4:
The next rational root is $ x = \dfrac{ 3 }{ 4 } $
$$ \frac{ 16x^2-9}{ 4x-3} = 4x+3 $$Step 5:
To find the last zero, solve equation $ 4x+3 = 0 $
$$ \begin{aligned} 4x+3 & = 0 \\[1 em] 4 \cdot x & = -3 \\[1 em] x & = - \frac{ 3 }{ 4 } \end{aligned} $$