Find roots of polynomial $$ p(x) = 4x^4-109x^2+225 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 5\\[1 em]x_2 &= -5\\[1 em]x_3 &= \frac{ 3 }{ 2 }\\[1 em]x_4 &= -\frac{ 3 }{ 2 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 5 } $ is a root of polynomial $ 4x^4-109x^2+225 $.
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}{ 225 } $, with factors of 1, 3, 5, 9, 15, 25, 45, 75 and 225.
The leading coefficient is $ \color{red}{ 4 }$, with factors of 1, 2 and 4.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 225 }}{\text{ factors of 4 }} = \pm \dfrac{\text{ ( 1, 3, 5, 9, 15, 25, 45, 75, 225 ) }}{\text{ ( 1, 2, 4 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 45}{ 1} \pm \frac{ 75}{ 1} \pm \frac{ 225}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 9}{ 2} \pm \frac{ 15}{ 2} \pm \frac{ 25}{ 2} \pm \frac{ 45}{ 2} \pm \frac{ 75}{ 2} \pm \frac{ 225}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 9}{ 4} \pm \frac{ 15}{ 4} \pm \frac{ 25}{ 4} \pm \frac{ 45}{ 4} \pm \frac{ 75}{ 4} \pm \frac{ 225}{ 4}\\[ 1 em] \end{aligned} $$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( 5 \right) = 0 $ so $ x = 5 $ 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-5 }$
$$ \frac{ 4x^4-109x^2+225}{ x-5} = 4x^3+20x^2-9x-45 $$Step 2:
The next rational root is $ x = 5 $
$$ \frac{ 4x^4-109x^2+225}{ x-5} = 4x^3+20x^2-9x-45 $$Step 3:
The next rational root is $ x = -5 $
$$ \frac{ 4x^3+20x^2-9x-45}{ x+5} = 4x^2-9 $$Step 4:
The next rational root is $ x = \dfrac{ 3 }{ 2 } $
$$ \frac{ 4x^2-9}{ 2x-3} = 2x+3 $$Step 5:
To find the last zero, solve equation $ 2x+3 = 0 $
$$ \begin{aligned} 2x+3 & = 0 \\[1 em] 2 \cdot x & = -3 \\[1 em] x & = - \frac{ 3 }{ 2 } \end{aligned} $$