Find the partial fraction decomposition of the rational function. (-10 x^2+27 x-14)/((x-1)^3(x+2

Find the partial fraction decomposition of the rational...

Key Concepts

Setting Up Equations for Coefficients

The process of partial fraction decomposition involves expressing the original function as a sum of unknown coefficients over simpler denominators, then clearing the denominators and equating coefficients of like powers of x. This method results in a system of linear equations that, once solved, gives the values of the unknown coefficients in the decomposition.

Repeated Linear Factors

When the denominator of a rational function includes repeated linear factors, each occurrence of the factor contributes a separate term in the partial fraction expansion. For a factor raised to a power, such as (x - a)^n, the decomposition must include fractions with denominators (x - a), (x - a)^2, ..., up to (x - a)^n.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. Understanding their structure is important because it allows you to decompose them into partial fractions when the degree of the numerator is less than that of the denominator, making them simpler to integrate or analyze.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to express a complicated rational function as a sum of simpler fractions. This is especially useful in integration, differential equations, and Laplace transforms because it breaks down complex expressions into terms that are easier to work with.

Transcript

00:01 Okay, we've been given this fraction and we'd like to put into partial fraction form.

00:04 You can see the factors down here are going to be x plus two, some sort of a, plus b over x minus one, plus c over x minus one squared, plus d over x minus one cubed.

00:23 Okay, dokey.

00:24 So, expanding these out, we're going to have a lots of x minus 1 cubed plus b lots of x minus 1 squared plus c lots of oh, sorry, x minus i've almost i messed up there x minus 1 squared x plus 2 plus c lots of x minus minus 1 x plus 2 plus d lots of just x plus 2 okay so we know this is going to be equal to minus 10 x squared plus 27 x minus 14 okay so um if we would put say x as 1 this bit would be 0 this bit would be 0 this bit would be 0 all we'd have to deal with is our d and whatever this bit comes out as.

01:28 So, well, 1 squared times minus 10 is just going to be minus 10 plus 27 minus 14.

01:37 And this is going to be equal to 1 plus 2, which is 3, 3d.

01:42 Okay, so you've got plus 27 minus 24 there.

01:46 So that's going to leave us with 3.

01:49 It is equal to 3d.

01:50 Divide both sides of 3.

01:52 D is equal to 1.

01:55 And so let's say x is equal to minus two this bill will be zero here so that's going to render zero this bit will be zero here so i'll be end of zero etc etc so we'll be left with our a so we will have minus two minus one is to be minus three minus three cube will be minus 27 a and that's going to be equal to well minus two squared is four so we'll have minus 40 plus 27 lots of x so that's going to be minus two lots of 27 which is going to be minus 54 and then we'll just have minus 14 so we'll have minus 27 a is equal to minus 100 and eight yes divided that by minus 27 we should get a is equal to minus four i'm hoping 27 54 yeah yeah yeah okay and so let's see what we're left with now we know that our a is minus four so we're gonna have minus four lots of x minus 1 cubed and we're gonna have still plus r b x minus 1 square x plus 2 and still plus r c x minus 1 plus two and then finally we'll just have one lot of what i'd be was which was plus x plus two and we know that this is equal to again our minus x square um plus 27 x minus 14 okay dokey so let's just expand these out and see what we have for our units x x squared and x cubes you have to figure out nice and easy then so um let's just do this here x minus 1 x minus 1 x squared minus x minus x plus one so it's x squared minus two x plus one for this bit if you times that by x minus one again x squared minus two x plus one x minus one x cubed whoops x cubed minus two x squared plus x minus x squared plus 2x minus 1 so that'll be x cubed minus 3x squared minus 3x minus 1 so i think that's oh no and then for the b wait do we have any other x cubes come out we actually don't so we know we're going to have b x cubes here and we know we don't have any extra x cubes out front.

05:16 So we know our b is going to be equal to minus 1.

05:21 So the only thing we'll have to work out now is the c value...

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