Find roots of polynomial $$ p(x) = -10x^4+76x^3-150x^2-42x+234 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= -1.0717\\[1 em]x_3 &= 1.9619\\[1 em]x_4 &= 3.7098 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ -10x^4+76x^3-150x^2-42x+234 $.
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}{ 234 } $, with factors of 1, 2, 3, 6, 9, 13, 18, 26, 39, 78, 117 and 234.
The leading coefficient is $ \color{red}{ 10 }$, with factors of 1, 2, 5 and 10.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 234 }}{\text{ factors of 10 }} = \pm \dfrac{\text{ ( 1, 2, 3, 6, 9, 13, 18, 26, 39, 78, 117, 234 ) }}{\text{ ( 1, 2, 5, 10 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 13}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 26}{ 1} \pm \frac{ 39}{ 1} \pm \frac{ 78}{ 1} \pm \frac{ 117}{ 1} \pm \frac{ 234}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 9}{ 2} \pm \frac{ 13}{ 2} \pm \frac{ 18}{ 2} \pm \frac{ 26}{ 2} \pm \frac{ 39}{ 2} \pm \frac{ 78}{ 2} \pm \frac{ 117}{ 2} \pm \frac{ 234}{ 2} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 2}{ 5} \pm \frac{ 3}{ 5} \pm \frac{ 6}{ 5} \pm \frac{ 9}{ 5} \pm \frac{ 13}{ 5} \pm \frac{ 18}{ 5} \pm \frac{ 26}{ 5} \pm \frac{ 39}{ 5} \pm \frac{ 78}{ 5} \pm \frac{ 117}{ 5} \pm \frac{ 234}{ 5}\\[ 1 em] \pm \frac{ 1}{ 10} & \pm \frac{ 2}{ 10} & \pm \frac{ 3}{ 10} & \pm \frac{ 6}{ 10} & \pm \frac{ 9}{ 10} & \pm \frac{ 13}{ 10} & \pm \frac{ 18}{ 10} & \pm \frac{ 26}{ 10} & \pm \frac{ 39}{ 10} & \pm \frac{ 78}{ 10} & \pm \frac{ 117}{ 10} & \pm \frac{ 234}{ 10} ~~ \end{aligned} $$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( 3 \right) = 0 $ so $ x = 3 $ 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-3 }$
$$ \frac{ -10x^4+76x^3-150x^2-42x+234}{ x-3} = -10x^3+46x^2-12x-78 $$Step 2:
The next rational root is $ x = 3 $
$$ \frac{ -10x^4+76x^3-150x^2-42x+234}{ x-3} = -10x^3+46x^2-12x-78 $$Step 3:
Polynomial $ -10x^3+46x^2-12x-78 $ has no rational roots that can be found using Rational Root Test, so the roots were found using qubic formulas.