SOLUTION: Let f(x)=-x^3+3x^2-3x+1, and g(x) be f(x) divided by 1-x; solve for g(x) if 1-x is a factor of f(x). a. g(x)=x^4-4x^3+6x^2-4x+1 b. g(x)=x^3-3x^2+3x-1 c. g(x)=-x^2+2x-1 d. g

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Let f(x)=-x^3+3x^2-3x+1, and g(x) be f(x) divided by 1-x; solve for g(x) if 1-x is a factor of f(x). a. g(x)=x^4-4x^3+6x^2-4x+1 b. g(x)=x^3-3x^2+3x-1 c. g(x)=-x^2+2x-1 d. g      Log On


   

Question 154518: Let f(x)=-x^3+3x^2-3x+1, and g(x) be f(x) divided by 1-x; solve for g(x) if 1-x is a factor of f(x).
a. g(x)=x^4-4x^3+6x^2-4x+1
b. g(x)=x^3-3x^2+3x-1
c. g(x)=-x^2+2x-1
d. g(x)=x^2-2x+1
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Let f%28x%29=-x%5E3%2B3x%5E2-3x%2B1, and g%28x%29 be f%28x%29 divided by 1-x; solve for g%28x%29 if 1-x is a factor of f%28x%29.
a. g%28x%29=x%5E4-4x%5E3%2B6x%5E2-4x%2B1
b. g%28x%29=x%5E3-3x%5E2%2B3x-1
c. g%28x%29=-x%5E2%2B2x-1
d. g%28x%29=x%5E2-2x%2B1

Let's first factor f%28x%29: 

f%28x%29=-x%5E3%2B3x%5E2-3x%2B1 

Regroup the terms:

f%28x%29=1-x%5E3-3x%2B3x%5E2,

Factor the first two terms as the difference of cubes:
Factor out -3x out of the last two terms:

f%28x%29=%281-x%29%281%2Bx%2Bx%5E2%29-3x%281-x%29

Factor out %281-x%29

f%28x%29=%281-x%29%28%281%2Bx%2Bx%5E2%29-3x%29 

Remove the inner parentheses:

f%28x%29=%281-x%29%281%2Bx%2Bx%5E2-3x%29

Combine the two like terms:

f%28x%29=%281-x%29%281-2x%2Bx%5E2%29

Factor the second parentheses:

f%28x%29=%281-x%29%281-x%29%281-x%29

Write as a cube:

f%28x%29=%281-x%29%5E3

Now since

g(x) is f(x) divided by 1-x



So the correct choice is (d)

Edwin