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JH
Numerade Educator

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Problem 2 Easy 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 - 6}{x^2 + x - 6} $ (b) $ \dfrac{x^2}{x^2 + x + 6} $

Answer

(a) $\frac{A}{x+3}+\frac{B}{x-2}$
(b) $1+\frac{A x+B}{x^{2}+x+6}$

Discussion

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Video Transcript

Let's go ahead and write out the form of the partial fraction to composition for the functions given in part A and B. So looking at part, eh, I see that there's a quadratic in the denominator, so it really depends on whether or not this thing factors. So first we see if we can factor in, and this one does factor. So we have here X plus three X minus two. And then this is what the book would call case one. You have distinct linear factors in the denominator, and so the partial fraction will be of the form A over X plus three. Please be over X minus two and we're asked to not determine the numerical values of the coefficients so we can stop here. We don't need to say what Andy are equals you now for part B. Well, first of all, do narrator and denominator have equal degree, so we will need long division. So let's go ahead and set up. Also, at some point, we'LL need to know whether or not this denominator factors. So we should use the test. We should look at the discriminating ear B squared minus four a. C. or the A, B and the sea are coming from just a generic polynomial. So does the book mentions. The polynomial won't factor over the real numbers If B squared minus four, A. C is less than zero. So let's check this. In our case, a equals B equals one. These are the coefficients in front of the X Square in the X See equal six. So B squared minus four A. C. It's just one minus four times one time six and that's less than zero. So this thing will not fact there. So we'LL come back and use this fact later on. Now coming over here to the long Division X squared over X squared is one. Go ahead and multiply the one to the X Square plus X Plus six, and then we subtract and we get negative X minus six. So since tells us that we can write Part B as one, the question and then plus the remainder, which is negative X minus six over the original denominator. Now we would use partial fraction decomposition at this point, and we know that the denominator doesn't factor because we looked at the discriminatory B squared minus four a. C was negative. So this is what the book would call Case three. We havejust one distinct quadratic form that doesn't factor, and it's not repeated. It's on ly being raised in the first power here, so the partial fraction to composition should be of the form. What we have are one here, so let's write the one and then we have a X plus me. We need a linear term up there and then our original denominator. And there's no need to find out what a NPR because we weren't asked to do not determine the value of the coefficients. But due to a coincidence from the previous line on the left hand side, you could see that they should be minus one and B should be minus six. But the circle part on the bottom right? That will be my final answer, and that's in here