The zeroes of x²–2x –8 is: 1.(2,-4) 2.(4,-2) 3.(-2,-2) 4.(-4,-4)

1.	The zeroes of x²–2x –8 is: 1.(2,-4) 2.(4,-2) 3.(-2,-2) 4.(-4,-4)
Answer» Answer: $\Large{\underline{\underline{\bf{Option\:2:\:(<klux>4</klux>,-2)}}}}$ Step-by-step explanation: $\Large{\underline{\underline{\bf{Given:}}}}$ p(x) = x² - 2X - 8 $\Large{\underline{\underline{\bf{To\:Find:}}}}$ Zeros of the polynomial $\Large{\underline{\underline{\bf{Solution:}}}}$ → x² - 2x - 8 = 0 → By splitting the middle term method, x² - 4x + 2x - 8 = 0 → Taking the common factors out, x (x - 4) + 2 (x - 4) = 0 (x - 4) (x + 2) = 0 → Either x - 4 = 0 x = 4 → Or x + 2 = 0 x = -2 → Hence the zeros are 4 and -2 $\boxed{\bold{Zeros\:are\:4\:and\:-2}}$ → THEREFORE option 2 is correct. $\Large{\underline{\underline{\bf{Notes:}}}}$ → The zeros of a polynomial can be FOUND out by Using FACTOR theorem $x=\dfrac{-b\pm\sqrt{b^{2}-4ac } }{2a}$ Splitting the middle term Completing the square method

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$\Large{\underline{\underline{\bf{Option\:2:\:(<klux>4</klux>,-2)}}}}$