Find the partial fraction decomposition of 4x²+5x-9/(x³- 6x-9)

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Alright. Hello. Today we're going to be performing partial fraction decomposition on the given expression. So what we are given is seven x squared plus six X plus 23 over x cubed minus 18, X minus 35. Now, before we start a partial fraction decomposition, we always want to check to make sure that our denominator is fully factored. Currently our denominator can be factored using the rational 00. And what that states is we go ahead and take a look at the first term and the last term which will be labeled as Q and P. And we want to go ahead and list out the factors of the last term over the first term or in other words, this is going to be P over Q. And those factors are going to be one, 5, 7 and 35/1. So the possible zeros for the denominator is going to be plus or minus +157 and 35. Now we're gonna go ahead and perform synthetic division on the coefficients of our denominator in order to factor it, but we need to go ahead and pick one of the zeros that's going to produce a remainder of zero. And we can go ahead and pick five. So if we pick five, two br zero, let's go ahead and start by listing out the coefficients of our denominator. Notice that we're missing an X squared term from our denominator while we can go ahead and put a zero for its coefficient since we don't have one. So the coefficients is going to be the coefficient of x cubed term which is one, the coefficient of the x squared term which is zero coefficient of the X term, which is negative 18. And the constant at the end which is negative 35. We bring down the one and then one times five is going to give us five and we put that under 00 plus five is going to give us five and five times five is going to give us 25 which we put under negative 18, negative 18 plus 25 is going to give us seven and seven times five is going to give us positive 35 which we put under the next term and negative 35 plus 35 is going to give us zero for the remainder. And so the fully factored version of this is going to give us X squared plus five X plus seven. But keep in mind that we use the +05. So in this case x is equal to five but we want to go ahead and have the factor of X. That's going to give us five. So if we just add and subtract five to both sides of this equation, we get x minus five is equal to zero. And so the factor that produces five is going to be X -5. And so we have x minus five times X squared plus five, X plus seven. And that's gonna be the fully factored version of our denominator. So let's go ahead and rewrite our given expression with the new denominator. We have seven X squared plus six X plus 23 over x minus five times X squared plus five X plus seven. And now let's go ahead and perform the partial fraction decomposition. X -5 is a linear quantity as its leading coefficient has an exponent of one. So that's going to produce the partial fraction of a over X -5. And the quantity X squared plus five, X plus seven is a quadratic quantity as the leading coefficient has an exponent of two. And that's going to produce the partial fraction of B X plus C over X squared plus five X plus seven. And what we want to go ahead and do is go ahead and sell for missing coefficients which is a B and C. However, what we want to go ahead and do first is get rid of all these denominators and we can go ahead and do that by multiplying everything by the lowest common denominator. And the L C. D in this case is going to be x minus five times X squared plus five X plus seven. So let's just go ahead and write out what it would be like to multiply everything by the L c D. We have our original expression seven X squared plus six X plus over our new denominator Times The LC. D. Which is going to be X -5 times X squared plus five X plus seven over one. Equal to the first parcel fraction which is a over x minus five times X minus five times X squared plus five X plus seven over one. And that's gonna be plus the second partial fraction which is B. X plus C. Over X squared plus five X plus seven times the L. C. D. Which is going to be X minus five times X squared plus five X plus seven over one. Now multiplying the left hand side by the L. C. D. Is going to completely cancel out the denominator. Multiplying the first partial fraction by the L. C. D. Is going to cancel out the denominator but leave us with X squared plus five X plus seven in the numerator. And multiplying the second partial fraction by the L. C. D. Is going to cancel the denominator. Believe us with x minus five in the numerator. So what we have is seven X squared plus six X plus 23 is equal to a times X squared Plus five x plus seven plus the quantity B X plus C times X minus five. Now, what we want to go ahead and do is simplify the right hand side of this equation. And in order to do that we're gonna go ahead and distribute a into this quantity X squared plus five X plus seven. And we're going to foil bx plus C. With x minus five. Doing so is going to give us a X squared Plus five AX plus seven eight plus B, x squared -5, Plus CX -5 C. And what we want to go ahead and do now is group up all of our x square terms together, all of our X terms together. And all of our constants together. And that's going to give us a X squared plus B X squared plus five A x -5, plus C. X plus seven eight minus five C. And finally what we can go ahead and do is factor out an X square term for my first group and factor out an X term from our second group. And what we are left with is seven X squared plus six, X plus 23 is equal to X squared times A plus B plus x times five A minus five B plus C Plus Our Constance which is seven a minus five C. Okay, now, what we want to go ahead and do is use this equation to come up with the system of equations. And in order to do that, we're gonna go ahead and equate the coefficients of our x squared terms together, the coefficient of the X terms together. And the constants together. And what that's going to give us is the first equation which is going to be the constants of expert on the right hand side, which is a plus B equal to the constant of x square on the left hand side, which is seven. So we're going to have a plus B Is equal to seven. For a second equation, the constant of x. On the right hand side is five A minus five B plus C. And on the left hand side it's going to be six. So what we have is five A minus five, B plus C is equal to seven. And finally, for our last equation, we're just going to set our constants on both the right hand side and left hand side equal to each other. So we're going to have seven A -5. C is equal to 23 and this is going to be the system of equations that we need to use in order to solve for our constance. Now, at this point we should be comfortable with solving a system of equations and once you do solve the system of equations, you should get a is equal to four, B is equal to three And C is equal to one. And now you want to go ahead and take this and plug it into your original partial fractions. Now recall the partial fractions we came up with earlier was a over X -5 plus B. X plus C over X squared plus five X plus seven. So replacing A with four Is going to give us four Over X -5, replacing B with three is going to give us three X plus C. and replacing C with one is going to give us three X plus one over x squared plus five X plus seven. And this is going to be the complete partial fraction decomposition of our given expression. And that means that the answer to this problem is going to be be so I hope this video helps you in understanding how to perform the partial fraction decomposition, and I'll go ahead and see you all in the next video.

Find the partial fraction decomposition of 4x²+5x-9/(x³- 6x-9)

Key Concepts

Partial Fraction Decomposition

Polynomial Long Division

Factoring Polynomials

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