Question Video: Partial Fraction Decomposition | Nagwa Question Video: Partial Fraction Decomposition | Nagwa

Question Video: Partial Fraction Decomposition Mathematics

Find 𝐴 and 𝐡 such that 4/((π‘₯ + 8)(π‘₯ βˆ’ 2)) = (𝐴/(π‘₯ βˆ’ 2)) + (𝐡/(π‘₯ + 8)).

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Video Transcript

Find 𝐴 and 𝐡 such that four over π‘₯ plus eight multiplied by π‘₯ minus two is equal to 𝐴 over π‘₯ minus two plus 𝐡 over π‘₯ plus eight.

What we’ve been asked to find here is the partial fractions decomposition of the quotient on the left-hand side. We’re separating out into the sum of two fractions, which have linear denominators. In order to find the values of 𝐴 and 𝐡, we first want to write each of the fractions on the right-hand side with the same denominator as we have on the left. So to create a common denominator of π‘₯ plus eight multiplied by π‘₯ minus two, we need to multiply both the numerator and denominator of the first fraction by π‘₯ plus eight and the numerator and denominator of the second fraction by π‘₯ minus two. That’s whichever linear factor isn’t already included in the denominator.

As the two fractions on the right-hand side will now have the same denominator, we can add them together. So the right-hand side becomes 𝐴 multiplied by π‘₯ plus eight plus 𝐡 multiplied by π‘₯ minus two all over π‘₯ plus eight multiplied by π‘₯ minus two. Now, as the fractions on each side of this equation are equal to one another and they now have the same denominator, we can write a new equation equating the numerators only. We have that four is equal to 𝐴 multiplied by π‘₯ plus eight plus 𝐡 multiplied by π‘₯ minus two. And We can use this equation to find the values of 𝐴 and 𝐡.

Remember, this equation is true for all values of π‘₯. So we can choose values of π‘₯ which will eliminate either 𝐴 or 𝐡 in turn in order to find the value of the other. Firstly, we can choose π‘₯ to be equal to negative eight because this gives the equation four equals 𝐴 multiplied by negative eight plus eight plus 𝐡 multiplied by negative eight minus two. And of course, negative eight plus eight is simply zero. So we’ve eliminated the term in 𝐴. We’re left with the equation four equals negative 10𝐡 which we can solve by dividing both sides by negative 10. 𝐡 will be equal to negative four-10ths which simplifies to negative two-fifths.

The second value of π‘₯ we should chooses is the value that will make the second set of parentheses equal to zero. So we choose π‘₯ equals two, giving four equals 𝐴 multiplied by two plus eight plus 𝐡 multiplied by two minus two. But two minus two is intentionally equal to zero. We have, therefore, eliminated the 𝐡s and we have the equation four is equal to 10𝐴. We solved by dividing both sides by 10, giving 𝐴 equals four-10ths which simplifies to two-fifths. So by considering the partial fractions decomposition of four over π‘₯ plus eight multiplied by π‘₯ minus two, we’ve found that the value of 𝐴 is two-fifths and the value of 𝐡 is negative two-fifths.

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