$p(x) = x^{2} - 9$. Zeros are

$p(x) = x^{2} - 9$. Zeros are
  1. $3$ and $-3$
  2. $9$ and $-9$
  3. $3$ only
  4. $-3$ only

Solution

$(x-3)(x+3) = 0 \Rightarrow x = \pm 3$.

Asked in: MH-SSC-9

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