Question

x²-2x-8 find the zeroes of the following quadratic polynomial als and verify the relationship between the zeros and​

Answer 1

Step-by-step explanation:

x^2+2x-4x-8 =0

x(x+2)-4(x+2) =0

(x+2)(x-4)=0

x=-2,4

zeros of the equation

Answer 2

$\huge {\fcolorbox{white} {lime} {Answer:}}$

$\sf x^2-2x-8 \\\\\sf \longrightarrow x^2-4x+2x-8 \\\\\sf \longrightarrow x(x-4)+2(x-4) \\\\\sf \longrightarrow (x+2)(x-4) \\\\\sf two \: zeroes \: are \: - 2 \: and \: 4$

$\fcolorbox{white} {lime} {Verification:}$

$Sum~ of~ roots~ = ~-2+4=2 \\\\\sf </p><p>= \frac{-b} {a} = \frac{-(-2)}{1} =2 \\\\\sf </p><p>Product~ of~ zeroes~ =~ 4 \times - 2 = - 8 \\\\\sf </p><p>= \frac{c} {a} = \frac{-8}{1}~=~-8 </p><p>$

Hence, verified.....