Question

find the zeroes of polynomial p(x)=x²+1\6x-2

Answer 1

$\huge\mathcal{Answer \ :-}$

$\mathsf{Given \ polynomial \ :-}$

$p(x) = {x}^{2} + \frac{1}{6} x - 2$
$\mathsf{Taking \ LCM, \ we \ get}$

$p(x) = \frac{6 {x}^{2} + x - 12}{6}$
$\mathsf{To \ find \ zeroes, \ p(x) \ = \ 0.}$

•°•

$\frac{6 {x}^{2} + x - 12 }{6} = 0 \\ \\ 6 {x}^{2} + x - 12 = 0 \\ \\ 6 {x}^{2} + 9x - 8x - 12 = 0 \\ \\ 3x(2x + 3) - 4(2x + 3) = 0 \\ \\ (2x + 3)(3x - 4) = 0 \\ \\ 2x + 3 = 0$
$\mathsf{And}$

$3x - 4 = 0 \\ \\ x = \frac{4}{3}$

$\mathsf{And}$

$x = \frac{ - 3}{2}$

$\mathsf\red{The \ zeroes \ are \ 4/3 \ and \ -3/2.}$

$\huge\mathbb{Hope \ this \ helps.}$