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)

50- 250 is correct i did that one too

step-by-step explanation:

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl