Expand the quotients in Exercises 1–8 by partial fractions.

Question

(t⁴ + 9) / (t⁴ + 9t²)

Accepted Answer

Welcome back everyone. In this problem, we want to expand X4 + 16 divided by X4 + 16 X2 by partial fractions. Now, to expand our expression, let's first address the fact that this is an improper fraction, right, because notice here the degree of the numerator 4 is equal to the degree of the denominator 4. So is there some way for us to manipulate our polynomial that will help us to factor out our denominator to make it easier to set up our partial fraction decomposition? Well, let's take our fraction. OK. And Let us add and subtract 16 x 2 from our numerator. In that case, this will be x 4. Let me put this in a different color here, plus 16 x 2. OK, let me just take this out for now just to show you, uh, what we have in mind, minus 16 x2, OK, plus 16. Now why would we do that? How does that help? Well, notice here that if we take our terms X4 + 16 X2, OK, and we group them together. Then we could decompose our fraction to become X4 + 16 X2 divided by X4 + 16 X2, OK. Minus or sorry plus 16 x2 + 16 divided by x4 + 16 x2 and now this is good because our numerator cancels our denominator here and now we're left with 1 + -16 x2 + 16 divided by x4 + 16 x2. So we've started our partial uh our our partial fraction decomposition. Next, we want to factor the denominator of the remainder term. OK, this is our remainder here. So how can we do that? Well, notice here that in the denominator x 4 + 16 x 2. It equals we could rewrite it as x2 multiplied by x2 + 16, so the denominator consists of a repeated linear factor x2 and an irreducible quadratic factor x2 + 16. So now if we were to set up our partial fraction decomposition, then we could take 16 x2 + 16 divided by x2 multiplied by x2 + 16. And rewrite it as a divided by x plus b divided by x2 plus c x + D divided by x2 + 16. So this is our expansion. This is we've expanded our remaining term. All we need to do now is to figure out what A, B, C, and D are so we can put them as constants or or terms, OK. So how can we do that? Well, if we multiply both sides by x2 multiplied by X2 + 16, then we end up with -16 x2 + 16 being equal to Ax multiplied by x2 + 16. Plus b multiplied by x2 + 16 Plus c x + D multiplied by x2. And if we expand here, we'll get AX cubed plus 16 AX + BX2 + 16B plus C X cubed plus DX2. And if we try to group our coefficients, OK, depending based on our variables, then we'll end up with A + C multiplied by X cubed, because that's A X cubed plus C X cubed. Plus B + D. Multiplied by x2. Plus 16 A X, OK, plus 16 B. So this is what -16 x2 + 16 would be, and now if we compare the coefficients from both sides of the equation, we can use it to solve for our unknowns. So starting with our X cubed term, right? So let me make a note of what we're doing here, comparing coefficients. OK, starting with our X cubed term. Then on the, we have A plus C as the coefficient of x cubed on our right hand side, and that would be equal to 0 on our left hand side because there is no x cubed term, OK. Next for our X2 term. We have b + D being equal to -160. For our x term we have 16a being equal to 0, which would imply that a equals 0, and for our constant term. We have 16b being equal to 16, which means that B equals 1. Now if that's the case, then since we know the values of A and B, we can use it to solve for C and D, because from A plus C equals 0, if A equals 0, that means C would also be equal to 0. And if that's the case, then from B + D equals -16. And B + D equals -16. Since we know B equals 1, that means 1 + D will be 16, which means that D equals -16 minus 1, which is 17. So C equals 0, D equals -17, A equals 0, and B equals 1. Now we can substitute these constants back into the decomposition and combine with the result from the division. So this means then that x 4 + 16 divided by x 4 + 16 x 2. Will be equal to 1 + 0 divided by x, which is 0 + 1 divided by x2 plus 0 x minus 17 or 17 divided by x2 + 16, which we could rewrite as 16-17 divided by x2 + 16. So this is going to be x 4 + 16 divided by X4 + 16 x2, expanded using partial fractions. Thanks a lot for watching everyone. I hope this video helped.

Expand the quotients in Exercises 1–8 by partial fractions.(t⁴ + 9) / (t⁴ + 9t²)

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Expand the quotients in Exercises 1–8 by partial fractions.
(t⁴ + 9) / (t⁴ + 9t²)