IB Mathematics AA HL Flashcards- Partial fractions

1// We should know how to add fractions, even with algebraic expressions, as so:

\[
\frac{4}{x+1} + \frac{2}{x-2} = \frac{4(x-2) + 2(x+1)}{(x+1)(x-2)} = \frac{6x-6}{x^2 – x – 2}
\]

But, how do we do it in reverse, and why would we want to do that?

[q]

PROCESS// Let’s try to reverse the example above:

1. FACTOR: \(\frac{6x-6}{x^2 – x – 2}\) → \(\frac{6x-6}{(x+1)(x-2)}\)

2. SPLIT: \(\frac{A}{x+1} + \frac{B}{x-2}\) (where A & B are constants)

[a]

3. FIND A & B: \(A(x-2) + B(x+1)\) must equal \(6x-6\).

\[
A(x-2) + B(x+1) = 6x-6
\]

\[
Ax – 2A + Bx + B = 6x – 6
\]

\[
Ax + Bx + (-2A + B) = 6x – 6
\]

\[
A + B = 6
\]

\[
-2A + B = -6
\]

Solving these, we find \(A=4\), \(B=2\).

[q]

4. WRITE PARTIAL FRACTION:

\[
\frac{4}{x+1} + \frac{2}{x-2}
\]

E.G.1 // Write \(\frac{13x-18}{2x^2 – 5x – 3}\) in the form:

\[
\frac{A}{Cx+d} + \frac{B}{Mx+N}
\]

[a]

1. Factor: \(2x^2 – 5x – 3 = (2x+1)(x-3)\)

\[
\frac{13x-18}{2x^2 – 5x – 3} = \frac{13x-18}{(2x+1)(x-3)} = \frac{A}{2x+1} + \frac{B}{x-3}
\]

\[
A(x-3) + B(2x+1) = 13x-18
\]

\[
Ax – 3A + 2Bx + B = 13x – 18
\]

\[
(A + 2B)x + (-3A + B) = 13x – 18
\]

Solving, we get \(A=7\) and \(B=3\). Hence,

\[
\frac{7}{2x+1} + \frac{3}{x-3}
\]

[q]

USES → binomial expansion// As \(\frac{A}{Cx+d} = A(Cx+d)^{-1}\), we can use the binomial expansion methods to expand partial fractions:

E.G.2 // Expand \(\frac{4-5x}{(1+x)(2-x)}\) up to the \(x^2\) term:

[a]

1. Express in partial fractions:

\[
= \frac{A}{1+x} + \frac{B}{2-x} \quad \text{where} \quad A=3, B=-2
\]

i.e.,

\[
\frac{3}{1+x} – \frac{2}{2-x}
\]

[q]

Expand:

1. First, with \(\frac{3}{1+x}\):

\[
3(1-x+x^2 + \ldots) = 3 – 3x + 3x^2 + \ldots
\]

[a]

2. Now, \(\frac{2}{2-x} = 2(1-\frac{x}{2}+\frac{x^2}{4} + \ldots)\):

\[
Combine: 3(1+x+x^2 + \ldots) – 2(1 – x + \frac{x^2}{4} + \ldots)
\]

\[
= 2 – 3x + 3x^2 + \ldots
\]

[q]

USES → integrals// (You need to know topic 5 for this use). If you had to integrate the expression in E.G.2 above:

\[
\int \frac{4-5x}{(1+x)(2-x)} dx
\]

[a]

We don’t immediately have a technique, but changing it to \(\frac{3}{1+x} dx – \frac{2}{2-x} dx\), this can be done relatively easily:

\[
= 3\ln(1+x) + 2\ln(2-x) + c
\]

[x] Exit text