Answer :

Given, p(x) = x^{3} + 1

To find the zeros of p(x), consider p(x) = 0

∴ x^{3} + 1 = 0

∴ (x + 1) (x^{2} – x + 1) = 0 Using the identity (a + b)^{3} = (a + b) (a^{2} – ab + b^{2})

∴ x + 1 = 0 or x^{2} – x + 1 = 0

∴ x = –1 ; real zeros of x^{2} – x + 1 are not possible.

∴ –1 is the only zero of p(x).

Since –1 is real, number of real zeros is 1.

The given polynomial has degree 3, so it can have 3 zeros at most.

⇒ No. of zeros = 3 ; No. of real zeros = 1

For p(x) = x^{3} + 1, taking x = –2, –1, 0 and 2.

We obtain the following table:

Plotting these points on graph paper, we obtain the following figure:

It can be seen that the graph intersects the X-axis at (–1, 0), so –1 is the zero of p(x).

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