Find the partial fraction decomposition for each rational express...

Question

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (4x^2 - x - 15)/(x(x + 1)(x - 1))

Accepted Answer

Welcome back. I am so glad you're here. We are asked to decompose the given rational function into partial fractions. Our given rational function is in the numerator. We have three P squared plus four, P plus 28. That's all divided by our denominator which is P multiplied by the quantity of P plus two multiplied by the quantity of P minus two. Our answer choices are answer choice. A seven divided by P plus four divided by the quantity of P plus two plus six divided by the quantity of P minus two. Answer choice B negative four divided by P plus seven divided by the quantity of P plus two plus six divided by the quantity of P minus two. Answer choice C negative four divided by P plus six, divided by the quantity of P plus two plus seven divided by the quantity of P minus two and answer choice D negative seven divided by P plus four divided by the quantity of P plus two plus six divided by the quantity of P minus two. All right. The first thing we ask ourselves when doing partial fraction decomposition is whether we have a true um proper fraction. And we do that by seeing if the deno the degree, excuse me, the degree of the numerator is less than the degree of the denominator. The degree of our numerator here is two and the degree of the denominator, we have a P multiplied by A P multiplied by A P. So that's three and two is less than three. So we're good to go. The next thing we ask ourselves is whether our denominator is fully factored. And in this case, it is. So now we ask ourselves some questions about our denominators factors. The first question is whether those factors are distinct or repeating. And P, there's only one, just A P that's distinct P plus two. That's the only P plus two factor and P minus two. That's the only P minus two. All three of these are distinct factors. The next question we ask ourselves is are these factors linear or quadratic? And P that's to the first degree, it's linear P plus two is to the first degree, it's linear P minus two. Also to the first degree, all three of these factors are linear. Now, what we do with that information is we set this fraction equal to and we have three distinct factors in the denominator. So it will be equal to three fractions added together. Now, the denominators of those three fractions are going to be our three distinct factors from the denominator on the left hand side. So the first fractions denominator is P. The second fractions denominator is P plus two and the third fractions denominator is P minus two. Now, for the numerators, that's where that linear or quadratic question comes in for linear denominators, they are going to have constant terms as their numerators, we don't know what they are. So we use a capital A for the first one, we have a divided by P. We use a capital B for the second one, B divided by P plus two. Then we use a capital C for the third one, C divided by P minus two. And we have this all set up. Now it's just a matter of using algebra in order to solve for A B and C. So the first thing we do to solve for A B and C is we get rid of these fractions. We do that by multiplying everything here by the denominator on the left hand side, which is going to be P multiplied by the quantity of P plus two, multiplied by the quantity of P minus two. And so when we do that multiplication, multiplying P times P plus two times P minus two by the first fraction on the left hand side that cancels out with the denominator. And we just have the, the numerator which is three P squared plus four P plus 28 that is equal to. Now taking our first fraction on the right hand side, A divided by P, the P in the denominator cancels out that P, so that A is only being multiplied by the quantity of P plus two and the quantity of P minus two and we can bring down plus B now that P plus two and the denominator for the B fraction is going to cancel out that P plus two. So now that B is only multiplied by the P and the quantity of P minus two, now we can bring down plus C, the P minus two. And the denominator of the C fraction cancels out that P minus two. So the C is only multiplied by P and the quantity of P plus two. And now we've got this set up and because all three of these were distinct linear factors, what we get to do is really fun. We get to plug in the zeros of the denominator of our original fraction. And it will help us solve for A B and C pretty quickly. So the zeros for here, we set each of these factors equal to zero, P equals zero. Well, there's one factor is zero, P plus two equals zero. Subtracting two from both sides, we get P equals negative two, negative two is another zero. And likewise setting P minus two equal to zero, adding two to both sides, we get P equals a positive two. Those are our three zeros for our denominators. So now we're going to plug in those zeros and we'll solve for A B and C. So let's start off with the 00. We have three multiplied not by P but now by zero squared plus four, multiplied not by P, but by zero plus 28 equals a multiplied by the quantity of zero plus two multiplied by the quantity of zero minus two plus. Now this part's fun B multiplied not by P but by zero, multiplied by the quantity of zero minus two plus C not multiplied by P but by zero multiplied by the quantity of zero plus two and lots of stuff cancels out. Now this three multiplied by zero squared. It's nothing four, multiplied by zero is nothing. The left hand side is just 28 that equals a multiplied by zero plus two is positive 20 minus two is negative two and B multiplied by zero, multiplied by zero minus two. That's nothing. C multiplied by zero, multiplied by zero plus two is nothing. So now solving for a, well, A is multiplied by two multiplied by negative two. So that's a negative four. A equals 28 dividing both sides by negative four, we get a equals negative seven and there's one of them. So now we need to find B we do that by plugging in our second zero, the negative two. So we'll have three multiplied not by P but by negative two squared plus four multiplied not by P but by negative two plus 28. Equals a multiplied by not P but negative two plus two multiplied not by P but by negative two minus two plus B multiplied by negative two, multiplied by the quantity of negative two minus two plus C multiplied by negative two, multiplied by the quantity of negative two plus two. And on the left hand side, negative two squared is positive four and four multiplied by three is 12. So we have 12 and then four multiplied by negative two is negative eight. So 12 minus eight plus 28. And when we take 12 minus eight plus 28 we get a positive 32 which is equal to. Now we have these negative two plus two factors. That's zero, negative two plus two is zero. So the A is getting multiplied by zero, the A is gone. Likewise, the C is getting multiplied by the negative two plus two factor C is getting multiplied by zero. So C is gone, we just have the B to worry about. So in parentheses, negative two minus two is negative four. That's being multiplied by negative two, negative two, multiplied by negative four is positive eight. So we have eight, B is equal to 32 dividing both sides by eight. When we get B equals a positive four, there's our B now for C, we're plugging in two for P and we know that that's going to give us a s of zero and BS of zero because they both have the P minus two factors. So we can just write down the left hand side and the C part. So we've got three multiplied by a positive two squared plus four, multiplied by a positive two plus 28 equal to. Now, for the C portion of this, we have C multiplied by a positive two, multiplied by the quantity of a positive two plus two. Now simplifying on the left hand side, two, squared is 43, multiplied by four is 12, four, multiplied by two is eight. So we have 12 plus eight plus 28 this time and 12 plus eight plus 28 gives us 48 on the left hand side. And then we have two plus two in parentheses, which is four multiplied by 22, multiplied by four is eight. So we have eight C on the right, dividing both sides by eight, we have C equal to a positive six. So we've solved for A B and C and what do we do with that? We're going to plug those in to our original partial fraction decomposition and then we'll have our answers. So plugging in for a, we have a negative seven. So in the numerator is negative seven divided by P and then that's plus our B is four. So in the numerator, we have four divided by the denominator is P plus two, then plus our C term is six. So we have six divided by the denominator P minus two. And that is it. We look at our answer traces and this matches with answer trace. D well done. We'll catch you on the next one.

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (4x^2 - x - 15)/(x(x + 1)(x - 1))

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Factoring Polynomials

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