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 - 3x - 4)/(x^3 + x^2 - 2x)

Accepted Answer

Welcome back. I am so glad you're here. We are asked to decompose the given rational function in partial fractions. Our given rational function is in the numerator three T squared minus T plus three. That's divided by the denominator which is T cubed minus two T squared minus three T. Our answer choices are answer choice. A one divided by T minus seven divided by four multiplied by the quantity of T plus one plus nine divided by four multiplied by the quantity of T minus three. Answer choice B negative one divided by T plus nine divided by four, multiplied by the quantity of T plus one plus seven divided by four multiplied by the quantity of T minus three. Answer choice C one divided by T plus seven, divided by four, multiplied by the quantity of T plus one plus nine, divided by four, multiplied by the quantity of T minus three. And answer choice D negative one divided by T plus seven, divided by four, multiplied by the quantity of T plus one plus nine divided by four multiplied by the quantity of T minus three. All right. So the first thing we ask ourselves when we are decomposing into partial fractions is if we have a proper fraction. And we see that by looking at the degrees of both our numerator and our denominator, the degree of the numerator here is two. The degree of the denominator is three. We need the degree of the numerator to be less than that of the denominator. And two is less than three. So this one works, we may proceed. The next step is that we are going to factor our denominator if it's not already factored. So we can copy over the numerator three T squared minus T plus three. That's all divided by. And in the denominator, we can factor out A T, all three terms have A T and then we'd be left with T multiplied by the quantity of T squared minus two T minus three. And we can further factor that T squared minus two, T minus three. I'm going to do it quickly. That would factor out to be T minus three multiplied by the quantity of T plus one. And so we've got that factored. And now what we're going to do is take a look at all of the factors in the denominator. And we're going to ask ourselves two questions first. Are they distinct or repeating the T that's distinct. There's only one of that factor T minus three, also distinct, only one of that factor T plus one, also distinct, only one of that factor. So they're all here distinct. Next, we ask ourselves are these linear or quadratic and all three of those factors are to the first degree. So all three of them are linear. So recalling from previous lessons, when we have distinct linear factors in the denominator, what we do is we set our fraction equal to because we have three factors. There are going to be three new fractions. The denominator for each of those three fractions is going to be each one, each of our factors. So the first fractions denominator is just T that factor. The second fractions denominator is T minus three. And the third fractions denominator is the third factor T plus one. And now because they are all linear, their numerators are going to be constant terms, but we don't know what they are. So we label them a capital A, A capital B and A capital C. And now all the rest of this is we're just solving for A B and C. So if you've got it set up like this, you're on the right track. Now, solving for A B and C. First thing we do is we are going to eliminate our fractions. We do that by multiplying them by the denominators that everything has, which is the denominators from the left hand side of the equal sign, multiplying everything by T multiplied by the quantity of T minus three multiplied by the quantity of T plus one. And for the fraction on the left-hand side of the equal sign, all of the denominators factors cancel out everything. It's being multiplied by. So we're left with just the numerator three T squared minus T plus three. And that's equal to. Now, we have the A, from the first fraction, the T factors cancel out and the A is multiplied by the factors T minus three and T plus one and plus the B for the B fraction, the T minus three factors cancel out. So the B is multiplied by the T and the quantity of T plus one then plus our C for the C fraction, the T plus ones cancel out. And the C is multiplied by the T and the factor T minus three. And we've got this set up now because all of these are linear. What we can do is find our zeros from our original fraction, the zeros of the denominator and then plug those zeros in for T. And that will help us solve for A B and C. So solving for our zeros for this T that would be A T equals zero. So one of our zeros is zero for T minus three. If we set that equal to zero and then solve for T by adding three to both sides, we get a zero of a positive three. And for T plus one, we set that equals zero, subtract one from both sides to solve. For T, we get a zero of negative one. So there's our third zero. Now, we can plug that in for T and solve and see what we get. So we have a three multiplied not by T squared, but now zero squared minus not T but zero plus three equals a multiplied by the quantity of not T but zero minus three multiplied by the quantity of not T but zero plus one plus B multiplied not by T but zero multiplied by the quantity of not T zero plus one plus C multiplied not by T but by zero, multiplied by the quantity of not T but zero minus three. And now because we've got a lot of multiplying by zeros. A lot of things just cancel out. This zero squared is zero, multiplied by three is zero. The minus zero is nothing. And on the left hand side, we just have three. That's equal to the A is being multiplied by zero minus three and zero minus three is a negative three and zero plus one is a positive one. So we have a multiplied by a negative three, multiplied by one which is negative three. So we have negative three A, the B is multiplied by zero, multiplied by one. So that's just zero. So there's no BC is multiplied by zero. So that's zero. So we just have three equals negative three. A solving for A dividing both sides by negative three, we get A equals negative one. So we've solved for our A great. Now, we can solve for B or C by plugging in our next zero, which is a re so we have three multiplied by three squared instead of T minus, instead of T three plus three equals a multiplied by not T but three minus three, multiplied by the quantity of not T but three plus one plus B multiplied by the re multiplied by the quantity of three plus one plus C multiplied by three, multiplied by the quantity of three minus three. And again, A, we're going to create a lot of zeros on the left hand side that we can do that work. Three squared is nine, multiplied by three is 27. And then that negative three plus three is zero. So this is 27 on the left hand side equal to, for the A, that three minus three, that's zero. So the A term is gone for B, we have B multiplied by three, multiplied by three plus 13 plus one is 43, multiplied by four is 12. So we have 12 B and then the C is multiplied by three, but it's multiplied by three minus three, which is zero. So the C term is gone and we can divide both sides by 12. We have B equals 27 12th. Those are both divisible by three. So that reduces to nine fourths, B equals nine fourths. All right. Now, let's solve for C we know that's the one left and that's the one we'll be looking at, we'll use that zero of negative one. So we have three multiplied by negative one squared minus negative one plus three on the left equal two. We know that our A and our B are going to zero out because they have that factor of T plus one. Plugging a negative one for T is going to be zero. So let's just look at the C, so we've got C multiplied by negative one, multiplied by the quantity of negative one minus three. And now simplifying on the left negative one squared is one. So we have three multiplied by one which is three, we have a three minus negative one which is three plus one plus three. So on the left three plus one plus three, that's seven equals on the right negative one minus three in the parentheses is negative four. That's being multiplied by negative one, which is going to be a positive four. So we have four C divide both sides by four and we get C equals seven fourths. And we have solved for A B and C and what do we do with that? Well, we take A B and C and plug those back in to our partial fraction decomposition for A B and C. And so plugging in for a, we have a negative one. So that will be a negative one divided by T plus our B is nine fourths. So that will be a nine divided by four, multiplied by the quantity of T minus three plus our C is +74. So that will be seven divided by the quantity of four multiplied by the quantity of T plus one. And that's it. Now, if we look at the answer choices, they have them in a different order, that's just because they put the factors in a different order when they set it up. So that's fine. As long as you have those same three fractions with their positives and negatives, they can be in a different order. We look at our answer choices and this matches with answer choice. 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 - 3x - 4)/(x^3 + x^2 - 2x)

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Factoring Polynomials

Watch next