F(x) = x3 -3x2 + 2x - 6 What is the real solution of f(x) ? Verify algebraically

Real solution is x = 3

complex zeros are -√2i and √2i

Product of linear factors;

(x-3)(x + √2i)(x-√2i)

Step-by-step explanation:

Here, we want to find the real solution of the cubic equation

We start by setting f(x) = 0 and factorizing it.

We have ;

x^3 - 3x^2 + 2x - 6 = 0

x^2(x-3) + 2(x-3) = 0

(x^2 + 2)(x-3) = 0

The real solution will be

x - 3 = 0

and this mean x = 3

To check if this is a zero of f(x), when we input x = 3 into the equation, we should get zero

Thus, we have;

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

27 -27 + 6 - 6

Adding all this equal zero and thus, x-3 is a linear factor of the function

We want to find the complex zeros;

set x^2 + 2 = 0

x^2 = -2

x = √-2

x = √2 * √-1

but √-1 is i

Thus, x = √2i

since square root can be positive of negative, we have that ;

x = √2i or x = -√2i

So the complex factors are;

x + √2i and x - √2i

So as a product of linear factors, we have;

(x-3)(x + √2i)(x-√2i)

Heres how to solve it
an infinite  geometric series  is the sum of an infinite  geometric sequence. this series would have no last term. the general form of the infinite geometric series is  a1+a1r+a1r2+a1r3++a1r+a1r2+a1r3+  , where  a1a1  is the first term and  rr   is the common ratio