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Express π₯ squared minus two over π₯ plus two times π₯ plus one squared in partial fractions.
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We have an algebraic fraction where the degree of the numerator is less than the degree of the denominator.
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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.
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And so we can decompose this into a partial fractions straightaway.
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We look at the denominator of this fraction which helpfully has already been factored for us.
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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.
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And the numerator of this fraction is some constant that weβll call π΄ for the moment.
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And the other factor is a repeated factor; itβs π₯ plus one squared.
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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.
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This is the form of the partial fraction decomposition of this fraction where π΄, π΅, and πΆ are constants that weβre going to have defined.
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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.
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On the left-hand side, we just have π₯ squared minus two.
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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.
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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.
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The π₯ plus two in the numerator of the first fraction cancels with its denominator.
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In the second fraction, we cancel one of the powers of π₯ plus one in the numerator with the denominator.
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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.
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And now we can tidy up.
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On the right-hand side, we can see that we just have a polynomial: π΄ times π₯ plus one squared plus π΅ times π₯ plus two times π₯ plus one plus πΆ times π₯ plus two.
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We can expand the parentheses in each term on the right-hand side.
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So for example, π΄ times π₯ plus one squared becomes π΄ times π₯ squared plus two π΄π₯ plus π΄.
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And we do that with the other terms too.
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And now we can combine lots of like terms on the right-hand side.
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We have an π΄π₯ squared and a π΅π₯ squared which we combine to get π΄ plus π΅ π₯ squared.
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Also terms of two π΄π₯, three π΅π₯, and πΆπ₯ which we combine to get two π΄ plus three π΅ plus πΆ times π₯.
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And finally, we get the constants term π΄ plus two π΅ plus two πΆ.
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The left-hand side is equal to the right-hand side for all values of π₯.
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And so the coefficients of π₯ squared on the left- and right-hand sides must be the same.
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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.
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On the right-hand side, the coefficient of π₯ squared is π΄ plus π΅.
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And on the left side, itβs just one.
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And as we said, these coefficients must be equal.
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So π΄ plus π΅ equals one.
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On the right-hand side, the coefficient of π₯ is two π΄ plus three π΅ plus πΆ.
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And on the left-hand side, there is no π₯ term, which is equivalent to there being a coefficient of zero.
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And so two π΄ plus three π΅ plus πΆ must equal zero.
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And finally, we compare the constants term.
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On the left, itβs negative two and on the right itβs π΄ plus two π΅ plus two πΆ.
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Now we have three linear equations in three unknowns: π΄, π΅, and πΆ.
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And we can solve these equations using our favorite methods to find the values of π΄, π΅, and πΆ.
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But there is another method which involves going back a few steps to the equation π₯ squared minus two equals π΄ times π₯ plus one squared plus π΅ times π₯ plus two times π₯ plus one plus πΆ times π₯ plus two.
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This is more than an equation; itβs an identity.
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It holds for all values of π₯.
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Itβs not π₯ we have to find from this equation, itβs the values of π΄, π΅, and πΆ which make this equation true for values of π₯.
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And if it is supposed to be true for all values of π₯, then certainly it must be true for π₯ equals negative one.
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So we can substitute that in.
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On the left-hand side, we get negative one squared minus two, which is negative one.
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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.
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And so the whole term is just going to evaluate to zero.
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So weβre only left with the term πΆ times π₯ plus two.
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And substituting in negative one to that, as negative one plus two is just one, we just get πΆ.
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So having the substituted π₯ equals negative one, we get that negative one is equal to πΆ, and hence πΆ is equal to negative one.
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We still have to find the values of π΄ and π΅.
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Can you see which value of π₯ we should substitute in to obviate some of the terms on the right-hand side?
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Well, we can substitute π₯ equals negative two.
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On the left-hand side, we get two.
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π΅ times π₯ plus two times π₯ plus one becomes π΅ times zero times negative one which is just zero.
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And similarly, πΆ times π₯ plus two after substitution becomes πΆ times zero which is just zero.
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And so the only term we have to worry about is π΄ times π₯ plus one squared which after substitution becomes π΄ times negative two plus one squared.
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And as negative two plus one is negative one and negative one squared is just one, we get that two equals π΄ or π΄ equals two.
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So what can we do to find π΅?
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We can substitute any number we like in now, so letβs take zero.
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On the left-hand side, we just get negative two.
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On the right-hand side, we get π΄ times one squared which is just π΄, plus π΅ times two times one which is two π΅, plus πΆ times two which is two πΆ.
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And you might think that weβre still going to have to find two more equations and solve them simultaneously.
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But in fact, we already know the values of π΄ and πΆ and we can substitute those in.
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So negative two is two plus two π΅ times [plus] two times negative one.
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This is just a linear equation in one variable, which we can solve to get π΅ equals negative one.
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We have now found the values of our constants π΄, π΅, and πΆ.
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And so weβre ready to write down our answer.
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This is: two over π₯ plus two plus negative one over π₯ plus one plus negative one over π₯ plus one squared.
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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.