Real solution is x = 3
complex zeros are -โ2i and โ2i
Product of linear factors;
(x-3)(x + โ2i)(x-โ2i)
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 = โ-2
x = โ2 * โ-1
but โ-1 is i
Thus, x = โ2i
since square root can be positive of negative, we have that ;
x = โ2i or x = -โ2i
So the complex factors are;
x + โ2i and x - โ2i
So as a product of linear factors, we have;
(x-3)(x + โ2i)(x-โ2i)