Question

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

Answer 1

Solution:

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

Let's group the expression

$x^3-3x^2+2x-6$

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

Now, take out common

$x(x^2+2)-3(x^2+2)$

$(x-3)(x^2+2)$

Now, we can see that,

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

The two factors which we gained for $p(x)=x^3-3x^2+2x-6$ are:-

1. (x-3) i.e. x = 3. This is first root of p(x).

2. x² + 2 => x² = -2 => x = ±√(-2) that means x = √(-2) and -√(-2)

Or it can be written as x = √(2i²) and -√(2i²)

For complex number we use i(iota) where i² = -1

So, the two roots are x = i√2 and -i√2

As p(x) is a cubic equation it was natural we will gain three roots(or zeroes) from it.

The three roots are => x = 3, x = i√2 and -i√2

And from these roots only x =3 is real root and other two roots are complex roots.

Thus, it gets proved that p(x)=x³ - 3x² + 2x - 6 has only one real zero​ that is x = 3.