Find the zeros of polynomial x²+1/6x-2

Let f(x) = x² + {tex} \frac 16{/tex}x - 2.
Then, f(x) = {tex} \frac 16{/tex}(6x²+ x -12) = {tex} \frac 16{/tex}(6x² + 9x - 8x - 12)
{tex} \Rightarrow{/tex} f(x) = {tex} \frac 16{/tex}{(6x² + 9x) - (8x + 12)}
= {tex} \frac 16{/tex}{3x (2x + 3) - 4 (2x + 3)}
= {tex} \frac 16{/tex}(2x + 3) (3x - 4)
Now f(x)=0 if

{tex}\style{font-family:Arial}{\style{font-size:12px}{\begin{array}{l}i.e.\;x=-\frac32\;or\;x=\frac43\\\end{array}}}{/tex}
Hence, {tex} ​​\alpha = \frac { - 3 } { 2 } \text { and } \beta = \frac { 4 } { 3 }{/tex} are the zeros of the given polynomial.
{tex}\style{font-family:Arial}{\style{font-size:12px}{\begin{array}{l}Now\;\alpha+\beta=\frac{-3}2+\frac43=\frac{-9+8}6=-\frac16\\\end{array}}}{/tex}

and {tex}\style{font-family:Arial}{\style{font-size:12px}{\begin{array}{l}\alpha\beta=\frac{-3}2\times\frac43=-2\\\end{array}}}{/tex}
The given polynomial is f(x) =x² + {tex} \frac 16{/tex}x - 2.
so a=1 b={tex}\frac16{/tex} c=-2

  {tex}\style{font-family:Arial}{\style{font-size:12px}{\begin{array}{l}\alpha+\beta=\frac ba=\frac{-{\displaystyle\frac16}}1=-\frac16\\\end{array}}}{/tex} 

and    {tex}\style{font-family:Arial}{\style{font-size:12px}{\begin{array}{l}\alpha\beta=\frac ca=\frac{-2}1=-2\\\end{array}}}{/tex}
Hence, the relation between the coefficients and zeros is verified.