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

Question

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

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 17. That's divided by the denominator which is T multiplied by the quantity of T plus one. Our answer choices are answer choice A 17 divided by T minus 17 divided by the quantity of T plus one answer choice B 17 divided by T plus 17 divided by the quantity of T plus one answer choice C negative 17 divided by T minus 17 divided by the quantity of T plus one and answer choice D negative 17 divided by T plus 17 divided by the quantity of T plus one. All right. The first thing we want to do when we are decomposing into partial fractions is see if we have a proper fraction, the degree of our numerator should be less than the degree of our denominator. And in the numerator, we have no ts whatsoever. So that's a degree of zero in the numerator. And in the denominator, we have two Ts being multiplied together. That's a degree of two and yes, zero is less than two. So we may proceed. And the second step, when we are, when we are decomposing into partial fractions is we want to make sure that our denominator is fully factored. And here T multiplied by T plus one that's fully factored. So the next thing that we do is we ask ourselves some questions about the factors in the denominator that contain variables. The first question we ask ourselves is, are these factors distinct or repeating? And looking at T there's only one single T factor and T plus one, that's the only T plus one factor. They are both distinct, neither of them repeats. The next question we ask ourselves about the factors in the denominator is, are they linear or quadratic? And this first T that's to the first degree. So that's linear. This T plus one, the T in there is to the first degree. So that is also linear. Both of these are distinct linear factors. So because we have two distinct factors, we set this fraction equal to two fractions being added together the denominators for those two fractions are the two factors from our original denominator. So the first fractions denominator will be T, the second fractions denominator will be T plus one. Now, for the numerators, that's where the linear part comes in. When we have a linear denominator, it's going to have a constant term for the numerator. And we don't know what it is. So we put a capital A which is divided by T. And then we put a capital B which is divided by the T plus one. And now all the rest of this work is just solving for A and B. Now, in order to do that, we're going to start by getting rid of the fractions, we get rid of the fractions by multiplying the whole thing by the denominator from our original fraction. So multiplying it all by T multiplied by the quantity of T plus one. And when we multiply that fraction on the left by T multiplied by T plus one, the denominator cancels out the T multiplied by T plus one. And we just have 17 on the left hand side. Now that's equal to, we take that A divided by T fraction, multiplying it by the T multiplied by T plus one, the T factors cancel out so that A is only multiplied by the quantity of T plus one. Now, for the B divided by T plus one fraction, the T plus one factor from the denominator cancels out that T plus one factor. So the B is only multiplied by the T. So we have plus BT now because it was only linear and distinct factors that we had. What we can do to solve for A and B is, we can plug in the zeros of our original denominator. So if we set both of these denominators equal to zero, T equals zero. Well, there's one of our zeros, it's zero and T plus one equal to zero will solve for T subtracting one from both sides and we get T equals negative one. There's another zero at negative one. We plug zero in and we'll be able to solve for a, we plug negative one in and we'll be able to solve for B. Thanks, linear distinct uh fractions. So now we've got the zero in for T. So we'll have bringing this down 17 equals a multiplied by the quantity of not T but zero plus one plus B multiplied not by T but by zero. Well, the B multiplied by zero is nothing inside the parentheses. Zero plus one is a positive one. So we have a multiplied by one equals 17. While that just gives us an A value of 17. Now we can do the same plugging in our other zero, negative one. So we have 17 equals a multiplied by the quantity of not T but negative one plus one plus B multiplied by the quantity of not T but negative one. Simplifying in the parentheses being multiplied by the A negative one plus one is zero. So that A is gone, we just have B multiplied by negative one which is a negative B equals 17. Solving for B by dividing both sides by negative one. We get B equals negative 17 and that's it. Now we just plug A and B in for A and B in our original partial fraction decomposition. So instead of a, in the numerator divided by T, it's 17, divided by T plus. Instead of B in the numerator, it's negative 17. So we can change that plus to a minus and then divided by T plus one. And that is it, that is our partial fraction decomposition. Nice work. We look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

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

Key Concepts

Partial Fraction Decomposition

Factoring the Denominator

Setting Up and Solving Equations for Coefficients

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