Question

show that p(x) = x^3 - 3x^2 + 2x- 6 has only one real zero​

Answer 1

Answer:

Supposing you mean:

p(x) = x^3 - 3x^2 + 2x - 6

This is a cubic equation, so it has three zeros by its nature.

Look at the graph or use the rational roots theorem to find the root at x = 3.

Use the factor theorem to convert this root into a factor:

(x - 3)

Divide by the known factor:

(x^3 - 3x^2 + 2x - 6) / (x - 3) = x^2 + 2

Use the zero product principle:

x^2 + 2 = 0

Subtract 2:

x^2 = -2

Take the square root:

x = ± i √2 where i^2 = -1

So what's been shown is that p(x) has only one REAL zero.