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### Course: Algebra (all content)>Unit 13

Lesson 13: Partial fraction expansion

# Partial fraction expansion

Sal performs partial fraction expansion upon (10x²+12x+20)/(x³-8). Created by Sal Khan.

## Want to join the conversation?

• Can someone explain the Bx+C idea?
• When decomposing into partial fractions, the numerator of each fraction needs to be less than the degree of the denominator. Thus, if the denominator is linear, then the numerator can only be constant. If the denominator is quadratic, then the numerator might be linear, or it might be constant. To allow for either possibility, you should use the most general linear expression: Bx + C. This allows for both possibilities, including the case in which B = 0 and the numerator is constant.
• At , Sal "gets rid of" the (Bx+C)(x-2) term by making x equal 2. But the original fraction is not defined when x is 2. Are we allowed to substitute values of x for which the original fraction is undefined?
• That's great that you were able to answer your own question!
(1 vote)
• I don't understand how he gets the Bx+C at .. even after listening to the explanation several times. Can anybody explain that to me please?
• Because you cannot simply separate (x-2) and the (x^2 + 2x + 4) into two fractions without having the extra Bx + C. This is because the two new fractions do not have the same denominator and must be multiplied in. Essentially what you're trying to do it separate a large fraction, factor out the denominator and break it into two or three smaller fractions. However, you must still be able to multiply them all back out to get the same fraction as in the beginning, hence the Bx+C/(x^2 + 2x + 4) is there to multiply with the A/(x-2) to get the original fraction. So you would cross multiply and end up with A(x^2 + 2x + 4) + (Bx+C)(x - 2) = 10x^2 + 12x + 20
• Can decomposition of cubic equations be like this:
(x^3±y^3) = (x±y)(x^2∓xy+y^2)
• I remember coming across a mnemonic for the sequence of signs in the factoring of the sum of cubes and the difference of cubes.

sum of cubes:
`a^3+b^3=(a+b)(a^2-ab+b^2)`
The sequence of the signs for the sum of cubes is:
(+) : + , - , +

difference of cubes:
`a^3-b^3=(a-b)(a^2+ab+b^2`
The sequence of the signs for the difference of cubes is:
(-) : - , + , +

If you notice,
the first sign in the formula is the Same as the sum or the difference,
the second sign in the formula is the Opposite of the sum or the difference,
That spells out the acronym SOAP (Same, Opposite, Always Positive) as our mnemonic.

I hope this enables you to remember which is which. This should help clean it up for you.

I previously offered this in “Difference of Cubes,” but I thought it would be appropriate here as well. Sorry for the repetition.
• How does "if 2 is a zero" that leads to x-2? I don't see how one automatically springs from the other.
• The way I understand it, it's best to see this as a consequence of the remainder theorem. If you've been following the algebra 2 playlist you should've come across that. Here's the relevant video: https://www.khanacademy.org/math/algebra2/polynomial_and_rational/dividing_polynomials/v/polynomial-remainder-theorem-to-determine-coefficient-example

Basically, here's the rule: something is a factor of a polynomial `p(x)` if dividing the polynomial by that something gives you no remainder (this is true not just for polynomials but anything, really!). The remainder theorem states that `(x-a)` is a factor of a polynomial `p(x)` if and only if `p(a) = 0` (0 being the remainder here). For our exercise, we know that, if we can figure out a value for `x` for which `x^3 - 8 = 0`, we'll have found our `a` and our corresponding factor `x-a`. You can pretty simply see that this value has to be 2. So, from the remainder theorem: `p(2) = 0 <=> x-2 is a factor of p(x)`.

Furthermore, remember that `x^3 - 8` is also a difference of cubes. In factoring we learnt a handy little shortcut formula for both the sums and differences of cubes. For the difference, we know that `a^3 - b^3 = (a - b)*(a^2 + ab + b^2)`. So you could've come up with Sal's answer, which he got through long division, simply by plugging numbers into that formula (provided you could remember it! :) ).

If anyone has any corrections or additions I'd love to hear them!
• Why does the numerator have to be a lesser degree than the denominator?
• At , he points out that he didn't factor the (x^2 + 2x + 4) any further because it ends up with imaginary numbers. But I can think of plenty of situations where one COULD factor something further. What happens then? The next video didn't seem to answer this question for me. In the next video, the rule demonstrated only seems to apply when there is more than one of the SAME thing (in that case, it's multiples of x-2). What about when there 3 or more different things? If, for example, I have x^3 - 4x in my denominator, could I make it into (x)(x-2)(x+2)? Would I still use A, B and C numerators?
• It would decompose into something of the form A/x + B/(x-2) + C/(x+2). The degree of the numerator has to be one less than the degree of the denominator.
• At 12;23, he says divide both sides by minus 1. Could someone please explain why it couldn't have remained -b=-3?
• because in the question you have B not -B If you were to but -b=-3 into the question you will have to change a few signs, so the best way is to put b=3 and directly substitute. hope that helped