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. 4/(2x^2 -5x -3)

Accepted Answer

Welcome back. I am so glad you're here. We are given the expression seven divided by three Y squared minus y minus four. And we are asked to express this in partial fraction decomposition? Well, we recall from previous lessons on partial fraction decomposition that we're going to take a look at the denominator. And first thing we notice is that this denominator is not yet factored. So we need to factor it. We'll split this into two sets of parentheses and we ask ourselves what times what gives us three Y squared. That's a three Y times A Y. And now we need two more numbers that when multiplied, give us negative four. But when combined with the three Y and the Y add up to a negative one. Y. So we can try out a minus four and a plus one. And then we can just check that really quick. First three Y times Y. Three Y squared outside's three Y times one is plus three Y insides negative four times Y is minus four Y. And lasts negative four times one is negative four combining our like terms in the middle, we have three y squared minus one Y minus four. Which, yes, that checks out. So we factor that correctly. We can set this equal to seven divided by the factors three, why minus four times Why? Plus one. And now we ask ourselves several questions about the denominators here. First, are these factors linear or quadratic? And those wise both have a degree of one. So they are linear factors Next are these factors distinct or are they repeating? Well, they are different from each other. They are distinct. And the last question which we've answered, how many factors are there? There are two of them. So because there are two are going to set this equal to two new fractions added together for the denominators for these two fractions, they are going to be the two factors that we identified. So the first fractions denominator is the first factor 3, -4. The second fractions denominator is the second factor. Why? Plus one. And because we've identified both of these factors as linear, then both of their numerator are going to be constant terms. The first fractions numerator is a capital A. And the second fractions numerator is a capital B. And now we have set that up for partial fraction decomposition. Now we're just going to go through and solve for A and B and then come back up here and fill those in. So to solve for A and B. First, I want to get rid of all three of these fractions and we do that by multiplying all three of them by the lowest common denominator, which is the denominator from the left hand side of the equal sign. So multiplying everything times three y minus four times Y plus one. So that will look like we will have seven times three Y minus four times Y plus one. All over three y - times y plus one equals capital A Times three Y - times y plus one. All over three, Y minus four plus capital B times three Y minus four times Y plus one. All over Y plus one. And now the best fun part here, any fraction that has a factor occurring in both the numerator and the denominator gets to have those factors cancel out. So on the left hand side of the equal sign, the three y minus four is in the Y plus ones, cancel out on the right hand side of the equal side. The first fraction three, y minus four is cancel out and the second fraction the y plus ones cancel out. Yes, those fractions are gone. Now this reads seven equals capital A times Y plus one plus capital B times three Y minus four. And now we can distribute those capital letters. So seven equals A, Y plus one, A. Or just A plus distributing the B, three, B, y minus four B. Now, I'd like to rewrite this grouping. Ry terms together and grouping are constant terms together. So this will be seven equals a, Y plus three B, Y plus a minus four B. And now, one more step for this section of the work, we're going to factor out that why? So we'll have seven equals Y times a plus three B Plus A -4 B. Okay, we recall from previous lessons on partial fraction decomposition that the coefficients of terms like terms on either side of the equal sign are going to be equal to each other. So the coefficient of why on the right hand side of the equation is a plus three B. And that is going to be equal to the coefficient of why? On the left hand side of the equation, but there is no Y over there. Well, the coefficient of that Y is zero, so this A plus three B. The coefficient of Y on the right hand side is equal to zero. Now we also know that the constant terms are equal to each other. So this a -4b. The constant term on the right hand side is equal to the constant term on the left hand side, which is seven. So now we've set up two equations and we can use this system of equations to solve for A and B. And then go back up and fill them in for our partial fraction decomposition. So let's solve for A and B. By multiplying the first equation by negative one and then we'll get our a terms to cancel out. So we'll have a negative one. A minus three B equals zero times negative one is still zero. Now we can add these two together, ace, cancel out negative four B minus three B is negative seven B. And that equals seven plus zero is seven. Now dividing both sides by -7, we've got B equals a negative one. Alright, we've solved for B. We can go back to the first equation and we can fill in a negative one for B. Up here. And so we can rewrite that equation. One will be A plus three times not B. But now negative one equals zero and that will be an A three times negative one. So A minus three equals zero. To solve for A. We can add three to both sides and we get A equals A positive three. All right. We have solved for A. And B. Now we can go back up to the top and fill those in to our partial fraction decomposition here. So instead of A. We're going to put three divided by three, y minus four. And instead of B that's going to be a negative one divided by y plus one. We look at our answer choices and this matches with answer choice B. Well done. We'll catch you on the next one.

In Exercises 9–42, write the partial fraction decomposition of each rational expression. 4/(2x^2 -5x -3)

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Factoring Polynomials

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