integral: partial fractions...?

Hi. I'm reviewing for my Calc II mid-term. I'm hoping for at least a 65% this time. :(

Anyway, I'm not quite sure about this review question from the section on partial fractions:

I guess I just can't remember how to factor that denominator. Could somebody point me in the right direction?

Re: integral: partial fractions...?

Well, the factors of 3 are 1, -1, 3 and -3. So you need to substitute these to see which will give a value of 0.

-1 works, so that means [x - (-1)], or rather (x + 1) is a factor. Long divide to get the quadratic factor.

Re: integral: partial fractions...?

I don't understand why you say "None will, so it does not factorise over the integers" and then immediately say "-1 works".

Re: integral: partial fractions...?

That is a typo, I had originally thought none worked, but then I realised -1 did. I thought I deleted what I had written previously, but it looks like I didn't. Editing now.

Re: integral: partial fractions...?

Check if the sum of the coefficients of a polynomial is 0, then x=1 is a root and ( x -1) a factor. If the sum of the coefficients of terms with odd power of variable is equal to the sum of the coefficients of terms with the even powers of the variable then x = -1 is a root and (x+1) a factor.

In this case the sum of the coefficients of x^3 and x is equal to 1+1=2. and the sum of the coefficient of x^2 and x^0 is -1+3 = 2.

Hence (x+1) is a factor. Now i am sure you know how to proceed further.

Re: integral: partial fractions...?

Thanks for the help. With x+1 as a factor, this led to

which is