Question

Find the value of the following p(x)=x^3-3x^2-2x+6 at x=root 2

Answer 1

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.

Answer 2

Answer : 0

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