Question

find the zeroes of the quadratic polynomial
${x}^{2} - 2x - 8$

Answer 1

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

$\mathsf{x^2 - 2x - 8}$

$\mathsf{=> \ x^2 - 4x + 2x - 8}$

$\mathsf{=> \ x (x - 4) + 2 (x - 4)}$

$\mathsf{=> \ (x + 2)(x - 4)}$

$\mathsf{=> \ (x + 2) \ = \ 0 \ and \ (x - 4) \ = \ 0}$

$\mathsf{=> \ x \ = \ -2 \ and \ x \ = \ 4}$

$\tt{The \ zeroes \ are \ -2 \ and \ 4.}$

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

Answer 2

$ANSWER$

plzz refers the attachment

HOPE IT HELPS