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Question
Find the zeroes of the polynomial `f(x) = x^2 ˗ 2x ˗ 8` and verify the relation between its zeroes and coefficients
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Solution
`x_2 ˗ 2x ˗ 8 = 0`
⇒ `x^2 ˗ 4x + 2x ˗ 8 = 0`
⇒ `x(x ˗ 4) + 2(x ˗ 4) = 0`
⇒ `(x ˗ 4) (x + 2) = 0`
⇒ `(x ˗ 4) = 0 or (x+2) = 0`
⇒ `x = 4 or x = −2`
Sum of zeroes =`4+(-2)=2=2/1="-(Coefficient of x)"/(("Coefficent of" x^2))`
Product of zeroes =`(4) (-2)=-8/1="(Constant term"/((Coefficcient of " x^2))`
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