Find all real values of x for which f(x) = 0. f(x) = x3 − x2 − 3x + 3

Question

 can you help me please i really need it?

Answer 1

This question is asking you for the zeros of the cubic polynomial. This means you have to factor it and set it equal to zero.

 

The way you factor the cubic polynomial is by by regrouping is by breaking the 4 terms into two parts:

 

x3 - x2     and   -3x + 3

 

Factor out what's common in each part

 

 

For x3 - x2, there's an x2 common in both terms.

 

For -3x + 3, there is a -3 common in both terms. 

 

 

After you factor out the commonalities, rewrite the expression

 

 

x3 - x2

x2 (x - 1)  <-- See how if I factor out a x2, what's left is (x-1) because if I distribute the x2, I am back where I started.

 

 

-3x + 3  

-3 (x - 1)  <--  Alternatively, I could have factored out a +3 instead of -3.  But this would result in 3 (-x -1), and both terms in parentheses must be the same.

 

 

You know you did this correctly if the terms in parentheses are the same. 

 

Combine these factors together:

 

(x-1) (x2 -3)  <-- the (x-1) comes from the term that is the same for both groups, the x2 comes from the first commonality you pulled out, and the-3 comes from the second one.

 

 

The result is (x-1) (x2 - 3) as the factored form.

 

 

You need to find the values of x when f(x) = 0, so go ahead and set each term = 0 and solve for x.

 

x-1 = 0  and x2 -3 = 0

Answer 2

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

 

 

x=1 is a solution, since 1^2 - 1^2 - 3(1) + 3 = 1 - 1 - 3 + 3 = 0 - 3 + 3 = -3 +3 = 0

 

Synthetic division:

 

 1 |     1    -1    -3      3

                 1     0     -3

    ____________________

          1    0     -3    0

 

x^2 - 3 = 0

 

x^2  = 3

 

x = +or- sqrt(3)

 

 

The solution set is { 1, +sqrt(3),  -sqrt(3) }