Question

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

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

The given polynomial has only one real zero x = 3.

$P(x) = x^3 - 3 x^2 + 2x - 6$

So x−3 is a factor and x=3 is a zero of above polynomial.

So, $P(x) = (x-3) (x^2 + ax + b)$

               $= x^3 + (a-3)x^2 + (b-3a)x - 3b$

Now comparing coefficients of $x^2$,

⇒ a − 3 = −3

⇒ a = 0

Comparing coefficient of $x^0$

⇒ −3b = −6

⇒ b = 2

∴  $P(x) = (x-3) (x^2 + 2) = 0$

i.e, $x = 3, +i\sqrt{2}, -i\sqrt{2}$

Thus given polynomial has only one real zero x = 3.

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