In Exercises 9–42, write the partial fraction decomposition of ea...

Question

In Exercises 9–42, write the partial fraction decomposition of each rational expression. 4x² - 7x - 3/(x^3 -x)

Accepted Answer

welcome back. I am so glad you're here. We are given the expression three A squared minus two A minus nine, divided by a cubed minus nine A were asked to express this in partial fraction decomposition. Alright. We recall from previous lessons on partial fraction decomposition that we're going to take a look at the denominator and this denominator is not yet factored. So that's the first step. We need to factor it. So copying over three A squared minus two a minus nine. And then our denominator, we can factor out in a so that will be eight times a squared minus nine. And we look at our a squared minus nine and we recognize that that's a difference of squares. So we can rewrite that. I'll copy down our numerator, three A squared minus two a minus nine over eight times. We can write that as two sets of parentheses. And we recall from previous lessons, we take the square root of both of these. So the square root of a squared is a. That goes in the first position in each set of parentheses, the square root of nine is three. That goes in the second position in each set of parentheses and then by definition one is plus one is minus and there we go. Now this is factored. So now we're going to ask ourselves three questions about these factors. The first question is are these factors linear or quadratic? And all three of those a's. They have a degree of one. So all three of those factors are linear. Our next question is are they distinct or are they repeating? And none of those repeat. So they are all distinct. Our last question is, how many factors are there? There are three of them. So we are going to set this equal to three separate fractions all added together. We'll work on the denominators first. The denominators for each of our fractions is going to come from the denominators on the left hand side. So the denominator for our first fraction is going to be that first factor A. The denominator for the second fraction is going to be the second factor a plus three. And the denominator for the third fraction is going to be our third factor A minus three. Now for our numerator are enumerators come from the fact that all three of these are linear because all three factors are linear. All three numerator are going to be constant variables. So for the first fraction are constant term. In the numerator is going to be a capital A. So a divided by A. And no, that's not one. They are separate ones capital and one is lower case for the second fraction. Are the numerator is going to be the constant term capital B. And for the third fraction, our numerator is going to be the constant term capital C. Okay, we've got that set up. Now, what we're going to do is solve for A. B and C. And then we're going to come back up here and plug them in. And then this will be our answer for partial fraction decomposition. Once we filled in that A B and C. But it is quite a journey and a bit of work to solve for A B and C. What we're going to do to start off is multiply everything here by our common denominator to get rid of our fraction and our common denominator comes from the left hand side of the equation. It's going to be that A times A plus three times a minus three. So I'll write what that looks like. We have the three A squared minus two A minus nine now times little A, times A plus three and times A minus three. All over a times a plus three times a minus three equal to capital A. Times are little A, times A plus three times a minus three. All over. Just a little A plus capital B, times little A times a plus three times a minus three. All over. Just a plus three plus our capital C, times little eight times a plus three times a minus three. All over a minus three. And now the most wonderful part here, we get to cancel out any factors that occur in both the numerator and denominator of a fraction. So on the left hand side we've got a little a and a plus three and an a -3. The first fraction on the right hand side, the little A's cancel out the middle fraction on the right hand side, the A plus three is cancel out. And the third fraction on the right hand side, the a minus three's cancel out. So now we can write down what we've got left, three A squared minus two A minus nine. On the left hand side of the equal sign that equals capital A times parentheses, A plus three times A minus three plus capital B. Times little A times a minus three Plus Capital C. Times Little A Times A Plus three. Okay, now the next step here is I would like to distribute these terms that are in parentheses and leaving our capital letters alone for a moment. So copying down the left hand side, three A squared minus two A minus nine equals capital A. Times. And we know from the work, we just did that A plus three times a minus three. Is that difference of squares? A squared minus nine plus capital B. Times distributing that little A. So little a times little A is a squared a times negative three is minus three A plus capital C times a times A is a squared and a times three is plus three A. Now we can distribute the capital letters, bring down the left hand side equals capital a little A squared for the eight times A squared then minus a nine capital A plus B. A squared -3, a big b plus big C A squared plus three. Little a big C. Okay, now what we want to do is regroup these. We want to group our A squared together and we want to group our little A's together and we want to group well we want to get our constant term by itself. So copying down the left hand side, we will get to it pretty soon. And with the a squares together, that's capital, A little A squared plus B, A squared plus C, A squared. Now our little A's minus three, A B plus three A. C. And then the minus nine A. Now that we've grouped them together, we can factor out our A squared and factor out our little A. So copying down three A squared minus two A minus nine. And sorry if my A's look like nines and my nines look like a S case factoring out the A squared a squared times a plus B plus C. Plus factoring out the little A. So little A times negative three big B plus three big C. And then bring down the -9 a. Okay, now we are recalling from those previous lessons on partial fraction decomposition that the coefficients of our like terms on either side of the equal sign are equal to each other. So the coefficient of a squared on the left is three and that is equal to the coefficient of a squared on the right, which is A plus B plus C. So we can write that as an equation. We can say that A plus B plus C. The coefficient of a squared on the right hand side of the equation is equal to three. The coefficient of a squared on the left hand side of the equation. So that's our first equation. Now for these for this little A. The coefficient on the right hand side of little A. Of the equation for little A. Is negative three B plus three C. And that's equal to the coefficient of little A. On the left hand side of the equation which is negative two. There's our second equation. Now for the third equation, the constant terms are equal to each other. So on the right hand side the constant term is negative nine. A. And that's equal to the constant term on the left hand side which is negative nine. That's our third equation. Yay, okay let's start off by solving for a that one's pretty close to done already. We'll divide both sides by negative nine and we will get capital A equals negative nine divided by negative nine. What happened there negative nine divided by negative nine equals one A equals one year. We've solved for one of them. I'm going to write that up here so we can see it when we scroll up A equals one. And now for this first equation we can rewrite it, filling in for A. So A is one is no longer A plus B plus C. but one plus B plus C equals three, we can subtract one from both sides. So that will give us a new equation. One of B plus C equals three minus one is two. And I'll copy over the second equation again because it has the same terms negative three, B plus three C equals negative two. And we can use these two equations to solve for B and C. Let's do some work with the first equation, let's multiply that times three and that will get RB terms to cancel out and we can solve for C. So three times B is three, B, three times C is plus three, C equals two times three is six. Now adding those two equations together, the bees cancel out three C plus three, C. Is six, C equals negative two plus six, which is four, dividing both sides by six. And we get c equals 2/3. Alright, We've solved for A And see just one more to go. Let's plug see into this first equation where we have B plus C equals two. So now that will be B plus not see but two thirds equals two. And we'll subtract 2/3 from both sides and that will give us B equals two minus two thirds is four thirds. So B equals four thirds. We have solved for A. B and C. Right now let's get back up here. Or you can see all our answer choices. I am going to plug A. B and C in here where we have our partial fraction decomposition set up. So instead of capital in the numerator here, this will be one Over little a. And instead of capital B in the numerator for our second fraction, that will be 4/3. And instead of capital C for the numerator for our third one, that will be two thirds. And now we want to look at our answer choices and we notice that they have all of their fractions with the a minus three in the second fraction position and the A plus three is in the third fraction position. So those just got switched around. But we look closely and our answer choices match with answer choice C. Well done. We'll catch on the next one.

In Exercises 9–42, write the partial fraction decomposition of each rational expression. 4x² - 7x - 3/(x^3 -x)

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