Video: Partial Fraction Decomposition

Express (π‘₯Β² βˆ’ 2)/(π‘₯ + 2)(π‘₯ + 1)Β² in partial fractions.

08:03

Video Transcript

Express π‘₯ squared minus two over π‘₯ plus two times π‘₯ plus one squared in partial fractions.

We have an algebraic fraction where the degree of the numerator is less than the degree of the denominator. So on the numerator we have a quadratic with degree two, and on the denominator we have a cubic, if we expanded things out, with degree three. And so we can decompose this into a partial fractions straightaway.

We look at the denominator of this fraction which helpfully has already been factored for us. The first factor, which is π‘₯ plus two, tells us that one of the fractions in the decomposition must have a denominator π‘₯ plus two. And the numerator of this fraction is some constant that we’ll call 𝐴 for the moment. And the other factor is a repeated factor; it’s π‘₯ plus one squared. And so we need two more fractions in our decomposition, one with denominator π‘₯ plus one squared and one just with denominator π‘₯ plus one. This is the form of the partial fraction decomposition of this fraction where 𝐴, 𝐡, and 𝐢 are constants that we’re going to have defined. And we find the values of these constants by multiplying both sides by the denominator of our original fraction, which is π‘₯ plus two times π‘₯ plus one squared.

On the left-hand side, we just have π‘₯ squared minus two. And on the right-hand side, the effect of multiplying by π‘₯ plus two times π‘₯ plus one squared is to just multiply each of those constants numerators by that. Having done that, we can see that each fraction on the right-hand side has a factor of the denominator in the numerator, and so we can cancel. The π‘₯ plus two in the numerator of the first fraction cancels with its denominator. In the second fraction, we cancel one of the powers of π‘₯ plus one in the numerator with the denominator. And finally, in the third fraction, we cancel the π‘₯ plus one squared in the numerator with the π‘₯ plus one squared which is the denominator.

And now we can tidy up. On the right-hand side, we can see that we just have a polynomial: 𝐴 times π‘₯ plus one squared plus 𝐡 times π‘₯ plus two times π‘₯ plus one plus 𝐢 times π‘₯ plus two. We can expand the parentheses in each term on the right-hand side. So for example, 𝐴 times π‘₯ plus one squared becomes 𝐴 times π‘₯ squared plus two 𝐴π‘₯ plus 𝐴. And we do that with the other terms too. And now we can combine lots of like terms on the right-hand side. We have an 𝐴π‘₯ squared and a 𝐡π‘₯ squared which we combine to get 𝐴 plus 𝐡 π‘₯ squared. Also terms of two 𝐴π‘₯, three 𝐡π‘₯, and 𝐢π‘₯ which we combine to get two 𝐴 plus three 𝐡 plus 𝐢 times π‘₯. And finally, we get the constants term 𝐴 plus two 𝐡 plus two 𝐢.

The left-hand side is equal to the right-hand side for all values of π‘₯. And so the coefficients of π‘₯ squared on the left- and right-hand sides must be the same. The coefficients of π‘₯ on the left- and right-hand sides must be the same and the constants term on left and right-hand sides must be the same. On the right-hand side, the coefficient of π‘₯ squared is 𝐴 plus 𝐡. And on the left side, it’s just one. And as we said, these coefficients must be equal. So 𝐴 plus 𝐡 equals one. On the right-hand side, the coefficient of π‘₯ is two 𝐴 plus three 𝐡 plus 𝐢. And on the left-hand side, there is no π‘₯ term, which is equivalent to there being a coefficient of zero. And so two 𝐴 plus three 𝐡 plus 𝐢 must equal zero. And finally, we compare the constants term. On the left, it’s negative two and on the right it’s 𝐴 plus two 𝐡 plus two 𝐢.

Now we have three linear equations in three unknowns: 𝐴, 𝐡, and 𝐢. And we can solve these equations using our favorite methods to find the values of 𝐴, 𝐡, and 𝐢. But there is another method which involves going back a few steps to the equation π‘₯ squared minus two equals 𝐴 times π‘₯ plus one squared plus 𝐡 times π‘₯ plus two times π‘₯ plus one plus 𝐢 times π‘₯ plus two. This is more than an equation; it’s an identity. It holds for all values of π‘₯. It’s not π‘₯ we have to find from this equation, it’s the values of 𝐴, 𝐡, and 𝐢 which make this equation true for values of π‘₯. And if it is supposed to be true for all values of π‘₯, then certainly it must be true for π‘₯ equals negative one. So we can substitute that in.

On the left-hand side, we get negative one squared minus two, which is negative one. Notice that on the right-hand side, as these two terms have a factor of π‘₯ plus one, when we substitute in negative one, these factors are going to become zero. And so the whole term is just going to evaluate to zero. So we’re only left with the term 𝐢 times π‘₯ plus two. And substituting in negative one to that, as negative one plus two is just one, we just get 𝐢. So having the substituted π‘₯ equals negative one, we get that negative one is equal to 𝐢, and hence 𝐢 is equal to negative one.

We still have to find the values of 𝐴 and 𝐡. Can you see which value of π‘₯ we should substitute in to obviate some of the terms on the right-hand side? Well, we can substitute π‘₯ equals negative two. On the left-hand side, we get two. 𝐡 times π‘₯ plus two times π‘₯ plus one becomes 𝐡 times zero times negative one which is just zero. And similarly, 𝐢 times π‘₯ plus two after substitution becomes 𝐢 times zero which is just zero. And so the only term we have to worry about is 𝐴 times π‘₯ plus one squared which after substitution becomes 𝐴 times negative two plus one squared. And as negative two plus one is negative one and negative one squared is just one, we get that two equals 𝐴 or 𝐴 equals two.

So what can we do to find 𝐡? We can substitute any number we like in now, so let’s take zero. On the left-hand side, we just get negative two. On the right-hand side, we get 𝐴 times one squared which is just 𝐴, plus 𝐡 times two times one which is two 𝐡, plus 𝐢 times two which is two 𝐢. And you might think that we’re still going to have to find two more equations and solve them simultaneously. But in fact, we already know the values of 𝐴 and 𝐢 and we can substitute those in. So negative two is two plus two 𝐡 times [plus] two times negative one. This is just a linear equation in one variable, which we can solve to get 𝐡 equals negative one.

We have now found the values of our constants 𝐴, 𝐡, and 𝐢. And so we’re ready to write down our answer. This is: two over π‘₯ plus two plus negative one over π‘₯ plus one plus negative one over π‘₯ plus one squared. And instead of writing plus negative one over something, we should write minus one over that something to get our final answer: two over π‘₯ plus two minus one over π‘₯ plus one minus one over π‘₯ plus one squared.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.