Partial Fractions - irreducibility question

[RJLiberator]

1. x^2-x+1

Is this factorable?

My initial thinking is NO. However, I can complete the square and it becomes (x-1/2)^2-3/4, but this doesn't seem to help me. Would this be considered factorable?

2. Turn 1/x^2-x+1 into partial fractions

Clearly, after I answer #1 correctly, #2 will be more clear. But under my initial assumption that #1 is NOT reducible, then how would I turn 1/x^2-x+1 into a partial fraction?
I am getting 1 = Ax+b where b = 0 and A = 1/x but that doesn't seem at all correct.

Thanks for any guidance / help.

 

[Mark44]

1. x^2-x+1

Is this factorable?

My initial thinking is NO. However, I can complete the square and it becomes (x-1/2)^2-3/4, but this doesn't seem to help me. Would this be considered factorable?

You have an error in your work. You should have (x - 1/2)² + 3/4, not (x - 1/2)² - 3/4. This quadratic is not factorable into factors with real number coefficients.

2. Turn 1/x^2-x+1 into partial fractions

This expression needs parentheses around the entire denominator. What you wrote is this:
$$\frac{1}{x^2} - x + 1$$

Clearly, after I answer #1 correctly, #2 will be more clear. But under my initial assumption that #1 is NOT reducible, then how would I turn 1/x^2-x+1 into a partial fraction?
I am getting 1 = Ax+b where b = 0 and A = 1/x but that doesn't seem at all correct.

No, it's not. The coefficients A and B should be constants, not expressions that involve a variable.

BTW, what happened to the homework template? You should not delete its parts.

 

[RJLiberator]

No, it's not. The coefficients A and B should be constants, not expressions that involve a variable.

Fair enough.

Do I use the denominator of x^2-x+1 or (x-1/2)^2 +3/4 to use partial fractions?

 

[LCKurtz]

Fair enough.

Do I use the denominator of x^2-x+1 or (x-1/2)^2 +3/4 to use partial fractions?

You don't turn ##\frac 1 {x^2-x+1}## into a partial fraction. In the second form above, where you have completed the square, it is already ready to integrate using an arctangent.

 

[RJLiberator]

LcKurtz, thank you my brother for the help verifying my thoughts.