Question

(1): The zeros of x2-9 are
(a) 3,-3
(b): -3,-3 (c): 3,-3 (d) none of these​

Answer 1

Answer:

Option (a) 3, - 3

Step-by-step explanation:

Let p(x) = x² - 9

For finding zeroes we equate p(x) to zero

p(x) = 0

x² - 9 = 0

x² = 9

x = 3, - 3

Option (a) 3, - 3

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Answer 2

Step-by-step explanation:

p(x)=x^2-9

p(x)=x^2-3^2

using identity a^2-b^2=(a+b)(a-b)

p(x)=(x+3)(x-3)

The zeroes of the polynomial are: 3 and- 3