Topic 1.11 · teacher page · AA Higher Level

One awkward curve is two simple ones

A decomposition with a purpose, not an algebra exercise.

The one thing to do with the animation

Point at the two asymptotes before you add anything.

The original blows up at x = 1 and x = -2. Each piece owns exactly one of those, which is how a student works out which factor a piece belongs to without being told.

Fading them in until they lie on the original makes the word decomposition mean something. This is not a rearrangement; it is the same function written as a sum.

Say why before you say how

1/(x - 1) integrates to a logarithm in one step. The original integrates to nothing in the form it arrives in. Partial fractions is a Topic 5 technique that happens to be taught in Topic 1.

Give that reason in the first two minutes or the whole thing looks like algebra for its own sake, which is exactly how it is usually received.

The answers

1. APut x = 2: 9 = 3A, so A = 3.
2. BPut x = -1: -6 = -3B, so B = 2.
3. Which x valuesB, the ones that make each bracket zero.

Where the marks go

1 markThe identity set up in the right form first.

1 markChoosing x to eliminate a bracket.

1 markA check at a third value.

What each wrong answer tells you

They giveWhat it means
2 for A (Q1)Swapped A and B. Worth asking which substitution each came from.
9 for A (Q1)Stopped before dividing by the bracket.
-2 for B (Q2)Sign slip: -6 over -3 is positive.
C (Q3)Will solve simultaneously, which works and is slower. The method exists to avoid it.

Other things they will say

"Can I substitute a value the original is undefined at?" Yes, because you are working with the multiplied-up identity, which holds everywhere. It is a fair bit of cheek and perfectly legitimate.

"What if the top is the same degree as the bottom?" Divide first. At this level it will not be, but knowing why the condition is there is worth thirty seconds.

"Does the order of A and B matter?" Only that each sits over the right factor. Checking at a third value catches a swap instantly.

A possible order

 What is happening
1The asymptotes, then fade the pieces in.
2Why: the integration that becomes possible.
3The cover-up method on two examples.
4Checking at a third value, every time.
5A question where the split is then integrated.

Two things not to say

Do not teach the method before the purpose. Without the integration it is unmotivated manipulation.

Do not skip the check. A sign error here is invisible and survives into the integration.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

  1. Split itExpress (3x + 5) / ((x + 1)(x + 3)) in partial fractions.
    1/(x + 1) + 2/(x + 3).
  2. Check your answerVerify that result at x = 0.
    1 + 2/3 = 5/3, and the original is 5/3 at x = 0.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

  1. Use the cover-upExplain how substituting x = −1 gives A immediately.
    At x = −1 the (x + 1) factor vanishes from the other term, leaving A = (3(−1) + 5) / (−1 + 3) = 1. Each root isolates its own numerator.
  2. Then integrateIntegrate (3x + 5) / ((x + 1)(x + 3)).
    ln|x + 1| + 2ln|x + 3| + c. Splitting the fraction is what makes it integrable at all.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

  1. Why the degree mattersExplain what must be done first if the numerator has degree equal to or greater than the denominator.
    Divide first. Partial fractions only apply to a proper fraction; otherwise the result is a polynomial plus a proper fraction, and skipping the division gives an answer that cannot be right.
  2. A repeated factorState the correct form for (2x + 1) / ((x + 1)²) and why.
    A/(x + 1) + B/(x + 1)². A repeated factor needs a term for each power, because two constants over the same linear factor cannot produce two independent degrees of freedom.

Practicalities

Works on a phone. Nothing is loaded from any other site, so it runs behind a school firewall, and nothing a student does is saved or sent anywhere.