Video: Partial Fraction Decomposition

Determine the partial fraction decomposition of (π‘₯Β² + π‘₯ + 1)/(π‘₯(π‘₯ βˆ’ 3)(π‘₯ + 1)Β²).

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Video Transcript

Determine the partial fraction decomposition of π‘₯ squared plus π‘₯ plus one over π‘₯ multiplied by π‘₯ minus three multiplied by π‘₯ plus one squared.

In order to answer this question, we need to recall how we write an expression of the form 𝑝π‘₯ squared plus π‘žπ‘₯ plus π‘Ÿ over π‘₯ multiplied by π‘₯ minus π‘Ž multiplied by π‘₯ minus 𝑏 squared in partial fractions. This can be written as four terms, 𝐴 over π‘₯ plus 𝐡 over π‘₯ minus π‘Ž plus 𝐢 over π‘₯ minus 𝑏 plus 𝐷 over π‘₯ minus 𝑏 squared. We can then multiply each of the five terms by the denominator on the left-hand side, π‘₯ multiplied by π‘₯ minus π‘Ž multiplied by π‘₯ minus 𝑏 squared. The left-hand side becomes 𝑝π‘₯ squared plus π‘žπ‘₯ plus π‘Ÿ. In the first term on the right-hand side, the π‘₯s cancel. So we’re left with 𝐴 multiplied by π‘₯ minus π‘Ž multiplied by π‘₯ minus 𝑏 squared.

In the second term, π‘₯ minus π‘Ž cancels. So we’re left with 𝐡π‘₯ multiplied by π‘₯ minus 𝑏 squared. In the third term, we can cancel one of the π‘₯ minus 𝑏s. This leaves us with 𝐢π‘₯ multiplied by π‘₯ minus π‘Ž multiplied by π‘₯ minus 𝑏. The final term becomes 𝐷π‘₯ multiplied by π‘₯ minus π‘Ž as the π‘₯ minus 𝑏 squared cancels. In this question, the two brackets or parentheses are π‘₯ minus three and π‘₯ plus one squared. This gives us π‘₯ squared plus π‘₯ plus one is equal to 𝐴 multiplied by π‘₯ minus three multiplied by π‘₯ plus one squared plus 𝐡π‘₯ multiplied by π‘₯ plus one squared plus 𝐢π‘₯ multiplied by π‘₯ minus three multiplied by π‘₯ plus one plus 𝐷π‘₯ multiplied by π‘₯ minus three.

There are two traditional methods for calculating the values of 𝐴, 𝐡, 𝐢, and 𝐷 from this point. One is by comparing coefficients. However, in this case, as there are so many terms, it will be easier to solve by substitution. The first two values that we will choose, three and negative one, are obvious, as these are the numbers that would make the parentheses zero. Negative one plus one is zero, and three minus three is equal to zero. The Third value we will choose is zero, as this would eliminate the second, third, and fourth term on the right-hand side. We could choose any value for our fourth substitution. In this case, we’ll choose one, as this will be easy to substitute.

Let’s begin by substituting π‘₯ equals three. On the left-hand side, we have three squared plus three plus one. On the right-hand side, the first, third, and fourth terms will become zero, as three minus three is zero and multiplying any number by zero is still zero. We are, therefore, left with 𝐡 multiplied by three multiplied by three plus one squared. Three squared is equal to nine. Three plus one is equal to four. Squaring this gives us 16 and multiplying by three gives us 48. Therefore, the right-hand side becomes 48𝐡. Adding nine, three, and one gives us 13. So 13 is equal to 48𝐡. Dividing both sides by 48 gives us a value of 𝐡 of 13 over 48.

Substituting in π‘₯ equals negative one gives us negative one squared plus negative one plus one on the left-hand side. On the right-hand side, the first three terms will cancel as negative one plus one is zero. The final term becomes 𝐷 multiplied by negative one multiplied by negative one minus three. Squaring negative one gives us positive one. And then adding negative one and adding one leaves us with one. The right-hand side simplifies to four 𝐷. Dividing both sides of this equation by four gives us a value of 𝐷 equal to one-quarter.

Our next step is to substitute π‘₯ equal zero. The left-hand side becomes one. As zero squared plus zero plus one is equal to one. The second, third, and fourth terms all become zero. This leaves us with 𝐴 multiplied by zero minus three multiplied by zero plus one squared on the right-hand side. Zero minus three is negative three. Multiplying this by one and then by 𝐴 gives us negative three 𝐴. Dividing both sides by negative three gives us a value for 𝐴 equal to negative one-third.

Our final substitution is π‘₯ equals one. On the left-hand side, we have three, as one squared plus one plus one is equal to three. The first term on the right becomes negative eight 𝐴. The second term is four 𝐡. The third term is negative four 𝐢. And the final term is negative two 𝐷. We can now substitute in the values we have already calculated for 𝐴, 𝐡, and 𝐷. Negative eight multiplied by negative one-third is equal to eight-thirds or eight over three. Four multiplied by 13 over 48 is 13 over 12. Negative two multiplied by one-quarter is negative one-half. The equation simplifies to three is equal to eight-thirds plus 13 over 12 minus four 𝐢 minus one-half.

Our next step is to group the constants on the right-hand side, eight-thirds plus 13 over 12 minus one-half. This is equal to 13 over four or 13 quarters. We can then subtract this from both sides of the equation, giving us negative one-quarter is equal to negative four 𝐢. If negative one-quarter is equal to negative four 𝐢, we can divide both sides by negative four. This gives us a value of 𝐢 equal to one 16th. We now have values for 𝐴, 𝐡, 𝐢, and 𝐷 of negative one-third, 13 over 48, one 16th, and one-quarter.

We can now write the initial expression as four partial fractions. The first term is negative one over three π‘₯. The second term is 13 over 48 multiplied by π‘₯ minus three. The third term is one over 16 multiplied by π‘₯ plus one. And the fourth and final term is one over four multiplied by π‘₯ plus one squared. This is the partial fraction decomposition of π‘₯ squared plus π‘₯ plus one over π‘₯ multiplied by π‘₯ minus three multiplied by π‘₯ plus one squared.

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