Video: Partial Fraction Decomposition

Express (π‘₯ βˆ’ 2)/π‘₯(π‘₯ βˆ’ 3) in partial fractions.

03:27

Video Transcript

Express π‘₯ minus two over π‘₯ times π‘₯ minus three in partial fractions.

We have the algebraic fraction π‘₯ minus two over π‘₯ times π‘₯ minus three, and we want to write it in partial fractions. That means decomposing it as the sum of two fractions, where the denominators of our two fractions are the factors of the original denominator.

We’re lucky that our denominator is already factored for us. We have π‘₯ times π‘₯ minus three, and so the denominators of our two fractions in our decomposition are going to be π‘₯ and π‘₯ minus three. The numerators of these two fractions are constants that we need to find. We’ll call them 𝐴 and 𝐡.

This is the form of our partial fraction decomposition. We just need to find the values of 𝐴 and 𝐡 now and we’ll be done, and we find the values of 𝐴 and 𝐡 by doing some algebra. We multiply both sides by π‘₯ times π‘₯ minus three, and we notice that we can cancel the denominator of each faction on the right-hand side with a factor of its numerator, so the π‘₯s cancel in the first fraction and the π‘₯ minus threes cancel in the second fraction.

So we are left with 𝐴 times π‘₯ minus three plus 𝐡π‘₯ on the right-hand side. We can expand out the brackets on the right-hand side β€” 𝐴 times π‘₯ minus three becomes 𝐴π‘₯ minus three 𝐴 β€” and then combine like terms, so 𝐴π‘₯ plus 𝐡π‘₯ becomes 𝐴 plus 𝐡 times π‘₯.

And as the equation we’re left with β€” π‘₯ minus two equals 𝐴 plus 𝐡 times π‘₯ minus three 𝐴 β€” must hold for all values of π‘₯, we can compare the coefficients. The coefficient of π‘₯ on the left-hand side is one and on the right-hand side is 𝐴 plus 𝐡, and these coefficients must be equal, so 𝐴 plus 𝐡 must be one.

And for the same reason, the constant term on the left-hand side, negative two, must be equal to the constant term on the right-hand side, negative three 𝐴. From this second equation, we can see that 𝐴 is two over three, and we can use this value in our first equation β€” 𝐴 plus 𝐡 equals one β€” to get that two-thirds plus 𝐡 equals one and hence that 𝐡 equals one-third.

Now that we have the values of 𝐴 and 𝐡, we can just put them into the expression we have. π‘₯ minus two over π‘₯ times π‘₯ minus three is two-thirds over π‘₯ plus one-third over π‘₯ minus three. And of course, we would prefer not to have a fraction in the numerator, so two-thirds over π‘₯ becomes two over three π‘₯ and a third over π‘₯ minus three becomes one over three times π‘₯ minus three.

And we get our final answer: π‘₯ minus two over π‘₯ times π‘₯ minus three expressed in partial fractions is two over three π‘₯ plus one over three times π‘₯ minus three. And you might like to check that this is true for all π‘₯s just by substituting different values of π‘₯ into this equation, not including π‘₯ equals zero or π‘₯ equals three when both sides of the equation are undefined.

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