F(x) = x3 -3x2 + 2x - 6 What is the real solution of f(x) ? Verify algebraically that your answer is a zero of f(x). Factor f(x). Find all complex zeros of f(x). Write f as a product of three linear factors.

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)

£56

step-by-step explanation:

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.