Find roots of polynomial $$ p(x) = 12x^4+11x^3+5x^2-7x-6 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -\frac{ 2 }{ 3 }\\[1 em]x_2 &= \frac{ 3 }{ 4 }\\[1 em]x_3 &= -\frac{ 1 }{ 2 }+\frac{\sqrt{ 3 }}{ 2 }i\\[1 em]x_4 &= -\frac{ 1 }{ 2 }- \frac{\sqrt{ 3 }}{ 2 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = -\dfrac{ 2 }{ 3 } } $ is a root of polynomial $ 12x^4+11x^3+5x^2-7x-6 $.
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}{ 6 } $, with factors of 1, 2, 3 and 6.
The leading coefficient is $ \color{red}{ 12 }$, with factors of 1, 2, 3, 4, 6 and 12.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 6 }}{\text{ factors of 12 }} = \pm \dfrac{\text{ ( 1, 2, 3, 6 ) }}{\text{ ( 1, 2, 3, 4, 6, 12 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 6}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 6}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 6}{ 3} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 6}{ 4} ~~ \pm \frac{ 1}{ 6} \pm \frac{ 2}{ 6} \pm \frac{ 3}{ 6} \pm \frac{ 6}{ 6} ~~ \pm \frac{ 1}{ 12} \pm \frac{ 2}{ 12} \pm \frac{ 3}{ 12} \pm \frac{ 6}{ 12} ~~ \end{aligned} $$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{ 2 }{ 3 } \right) = 0 $ so $ x = -\dfrac{ 2 }{ 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}{ 3x+2 }$
$$ \frac{ 12x^4+11x^3+5x^2-7x-6}{ 3x+2} = 4x^3+x^2+x-3 $$Step 2:
The next rational root is $ x = -\dfrac{ 2 }{ 3 } $
$$ \frac{ 12x^4+11x^3+5x^2-7x-6}{ 3x+2} = 4x^3+x^2+x-3 $$Step 3:
The next rational root is $ x = \dfrac{ 3 }{ 4 } $
$$ \frac{ 4x^3+x^2+x-3}{ 4x-3} = x^2+x+1 $$Step 4:
The solutions of $ x^2+x+1 = 0 $ are: $ x = -\dfrac{ 1 }{ 2 }+\dfrac{\sqrt{ 3 }}{ 2 }i ~ \text{and} ~ x = -\dfrac{ 1 }{ 2 }-\dfrac{\sqrt{ 3 }}{ 2 }i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.