Find the positive zeros of the quadratic polynomials $x^{2} - 2 x - 8$ .

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Solution

$x^{2} - 2 x - 8 = x^{2} - 4 x + 2 x - 8$
$= x (x - 4) + 2 (x - 4) = (x + 2) (x - 4)$
Zeroes are $- 2$ and $4$

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Question 1 (i)
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
$x^{2} - 2 x - 8$

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Find the positive zeros of the quadratic polynomials x2−2x−8.

x2−2x−8=x2−4x+2x−8=x(x−4)+2(x−4)=(x+2)(x−4)Zeroes are −2 and 4

Find the positive zeros of the quadratic polynomials $x^{2} - 2 x - 8$ .

$x^{2} - 2 x - 8 = x^{2} - 4 x + 2 x - 8$
$= x (x - 4) + 2 (x - 4) = (x + 2) (x - 4)$
Zeroes are $- 2$ and $4$