Find roots of polynomial $$ p(x) = 6x^4+53x^3-95x^2-25x+42 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \frac{ 3 }{ 2 }\\[1 em]x_2 &= \frac{ 2 }{ 3 }\\[1 em]x_3 &= -0.6782\\[1 em]x_4 &= -10.3218 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 3 }{ 2 } } $ is a root of polynomial $ 6x^4+53x^3-95x^2-25x+42 $.
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}{ 42 } $, with factors of 1, 2, 3, 6, 7, 14, 21 and 42.
The leading coefficient is $ \color{red}{ 6 }$, with factors of 1, 2, 3 and 6.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 42 }}{\text{ factors of 6 }} = \pm \dfrac{\text{ ( 1, 2, 3, 6, 7, 14, 21, 42 ) }}{\text{ ( 1, 2, 3, 6 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 7}{ 1} \pm \frac{ 14}{ 1} \pm \frac{ 21}{ 1} \pm \frac{ 42}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 7}{ 2} \pm \frac{ 14}{ 2} \pm \frac{ 21}{ 2} \pm \frac{ 42}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 6}{ 3} \pm \frac{ 7}{ 3} \pm \frac{ 14}{ 3} \pm \frac{ 21}{ 3} \pm \frac{ 42}{ 3} ~~ \pm \frac{ 1}{ 6} \pm \frac{ 2}{ 6} \pm \frac{ 3}{ 6} \pm \frac{ 6}{ 6} \pm \frac{ 7}{ 6} \pm \frac{ 14}{ 6} \pm \frac{ 21}{ 6} \pm \frac{ 42}{ 6}\\[ 1 em] \end{aligned} $$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( \dfrac{ 3 }{ 2 } \right) = 0 $ so $ x = \dfrac{ 3 }{ 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}{ 2x-3 }$
$$ \frac{ 6x^4+53x^3-95x^2-25x+42}{ 2x-3} = 3x^3+31x^2-x-14 $$Step 2:
The next rational root is $ x = \dfrac{ 3 }{ 2 } $
$$ \frac{ 6x^4+53x^3-95x^2-25x+42}{ 2x-3} = 3x^3+31x^2-x-14 $$Step 3:
The next rational root is $ x = \dfrac{ 2 }{ 3 } $
$$ \frac{ 3x^3+31x^2-x-14}{ 3x-2} = x^2+11x+7 $$Step 4:
The solutions of $ x^2+11x+7 = 0 $ are: $ x = -\dfrac{ 11 }{ 2 }-\dfrac{\sqrt{ 93 }}{ 2 } ~ \text{and} ~ x = -\dfrac{ 11 }{ 2 }+\dfrac{\sqrt{ 93 }}{ 2 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.