Problem 4 Medium Difficulty

Write out the form of the partial fraction decomposition of the function (as in Example 7). Do not determine the numerical values of the coefficients.

(a) $ \dfrac{x^4 - 2x^3 + x^2 + 2x - 1}{x^2 - 2x + 1} $ (b) $ \dfrac{x^2 - 1}{x^3 + x^2 + x} $

Answer

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Video Transcript

let's find the partial fraction decomposition for both of these functions. And there's no need to determine the numerical values of the coefficients. So looking at the first fraction, the first thing that I noticed is that the numerator is a higher degree, then the denominator. So any time the numerator has degree equal to or larger than the denominator, the first thing we should always do is long division, our polynomial division. And that's what we have set up over here on the right. So we'd liketo find the quotient here. So we have an X square exit, the forest. So we're gonna x squared up here, and then we multiply two x cubed plus X square, and then we subtract the whole thing doing so we have a lot of cancellation here, okay? And then we're just left with two X minus one. Now, this quadratic that we're dividing by does not go into to explain this one. So we have for part A. We can rewrite this as the question which is X square, and then we have the remainder over. The thing that we originally divided by this is why the long division is useful now every time we finished long division that they nominator will have larger degree than the numerator. And now we do partial fraction the composition. So as before, the first thing always look at that denominator. See if it can be factored here we can factor. This is X minus one squared. So this is what the book calls case, too. This's the repeated linear factor, which in this case, is just X minus one. So using the formula for case too x square, this's not a fraction less his write that as it is and then for this fraction we have a over X minus one plus be over. It's not this one square, and that will be our partial fraction. The composition for party. Let's go to the next page for party. So for me X squared minus one X cube X square in the next. So we noticed that the denominator has degree larger than the numerator. That's good. Not division necessary. Might be helpful to factor the numerator, but really more concerned with the denominator unless we could factor things out. So maybe I should go ahead and actually do this. Not always useful, but it can't be helpful. Sometimes you might be able to cancel things, though, but in this case, it doesn't look like that will happen in the denominator. Let's take out a X and there's our factories ation. This term, we can't factor. It's a linear. But then the second turn this may factor into two Lanier's So we need to look at the discriminative B squared minus four a. C Yeah, or again, thes coefficients are coming from just some polynomial this form. All right, In our problem, we see that a equals B equals C equals one finger. So that B squared minus four A. C is negative three, which is less than zero. And that tells us that this polynomial does not factor this guy appear. So that puts us in Case three. This is the case in which we just have a quadratic that doesn't factor not repeated. So this is what the text calling the textbook is calling Case three. No, For the first term, I see a X. So I just have a constant over X and then for the next term, because it's a quadratic on the bottom, it's not enough to have just a constant up top. You need a linear polynomial. So that's where I'Ll put the explosive and there's our answer for party.