Find the zeroes of the following quadratic polynomials $x^{2} - 2 x - 8$ .

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Solution

Comparing given polynomial with the general form $a x^{2} + b x + c$ ,

We get, a = 1, b = - 2 and c = -8

We have, $x^{2} - 2 x - 8$ = $x^{2} - 4 x + 2 x - 8$

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

= (x+2) (x-4)

Equating this to 0, we will get the values of 2 zeroes of the polynomial.

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

$\Rightarrow$ x = -2 and 4 are two zeroes.

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Find the zeroes of the following quadratic polynomials x2−2x−8.

Comparing given polynomial with the general form ax2+bx+c, We get, a = 1, b = - 2 and c = -8 We have, x2–2x−8 = x2–4x+2x–8 = x (x-4) +2 (x-4) = (x+2) (x-4) Equating this to 0, we will get the values of 2 zeroes of the polynomial. (x-4)(x+2) = 0 ⇒ x = -2 and 4 are two zeroes.

Find the zeroes of the following quadratic polynomials $x^{2} - 2 x - 8$ .