Mathematics (9709) | Algebra

Pure Mathematics 3: Algebra (Chapter 3.1)

Welcome to the Algebra chapter for Pure Mathematics 3 (P3)! This section is absolutely vital. While you might recognize some topics from P2, P3 deepens your understanding, introducing advanced techniques like Partial Fractions and the Generalized Binomial Theorem, which are cornerstone skills for later topics, especially complex integration and approximations.

Don't worry if the concepts look intimidating. We'll break them down step-by-step. Mastering these algebraic manipulations is key to unlocking marks in the more challenging P3 problems!

Key Topic 1: The Modulus Function (Absolute Value)

What is Modulus?

The modulus of a number, denoted by \( |x| \), is simply its distance from zero on the number line. Since distance cannot be negative, the output of the modulus function is always non-negative.
Example: \( |-5| = 5 \) and \( |5| = 5 \).

Solving Modulus Equations

When solving equations involving the modulus function, the safest and most reliable method is to square both sides.

Rule: \( |a| = |b| \implies a^2 = b^2 \)

This works because squaring a number automatically removes the negative sign, accounting for both possible cases simultaneously.

Example: Solve \( |3x - 2| = |x + 4| \).

1. Square both sides:
\( (3x - 2)^2 = (x + 4)^2 \)
2. Expand and simplify:
\( 9x^2 - 12x + 4 = x^2 + 8x + 16 \)
\( 8x^2 - 20x - 12 = 0 \)
3. Solve the quadratic (divide by 4):
\( 2x^2 - 5x - 3 = 0 \)
\( (2x + 1)(x - 3) = 0 \)
Solutions: \( x = -\frac{1}{2} \) or \( x = 3 \).

Solving Modulus Inequalities

Inequalities require careful handling, as simply squaring might complicate things if one side is a constant. We use two main rules based on whether the inequality is 'less than' or 'greater than'.

Rule 1: "Less Than" (The Sandwich Rule)
\( |x - a| < b \iff a - b < x < a + b \)
(Think of the solutions being "sandwiched" between two values.)

Rule 2: "Greater Than" (The Split Rule)
\( |x - a| > b \iff x - a > b \) OR \( x - a < -b \)
(Think of the solutions splitting off in opposite directions.)

Analogy: Imagine a speed camera (the modulus sign). If you are driving 'less than' the limit (Rule 1), you are safe between two speeds. If you are driving 'greater than' the limit (Rule 2), you are in trouble on the high side OR the low side!

Key Topic 2: Polynomials, Division, and Theorems

Polynomial Division

P3 expects you to be able to divide a polynomial (up to degree 4) by a linear or quadratic polynomial. This process is very similar to long division in arithmetic.

When you divide a polynomial \( P(x) \) by a divisor \( D(x) \), you get a Quotient \( Q(x) \) and a Remainder \( R(x) \):
\( P(x) = D(x) \cdot Q(x) + R(x) \)
The degree of the remainder \( R(x) \) must be less than the degree of the divisor \( D(x) \).

The Remainder Theorem

The Remainder Theorem is a fantastic shortcut!

Theorem: When a polynomial \( P(x) \) is divided by a linear factor \( (x - a) \), the remainder is \( P(a) \).

Important P3 Extension: If the divisor is of the form \( (ax + b) \), the value of \( x \) that makes the divisor zero is \( x = -b/a \). Therefore, the remainder is \( P\left(-\frac{b}{a}\right) \).

Example: Find the remainder when \( P(x) = x^3 - 4x + 1 \) is divided by \( (2x - 1) \).
We set \( 2x - 1 = 0 \), so \( x = 1/2 \).
Remainder \( = P(1/2) = (1/2)^3 - 4(1/2) + 1 \)
Remainder \( = 1/8 - 2 + 1 = 1/8 - 1 = -7/8 \).

The Factor Theorem

The Factor Theorem is just a special case of the Remainder Theorem.

Theorem: If the remainder \( P(a) = 0 \), then \( (x - a) \) is a factor of \( P(x) \).

We use this theorem extensively to find unknown coefficients in polynomials or to find the roots of polynomial equations.

Common Mistake: Forgetting to check the \( x \) coefficient! If \( (3x - 6) \) is a factor, then \( x = 2 \) must give a remainder of zero, not just that \( x=6 \) gives zero. Always find the value of \( x \) that makes the linear factor zero.

Key Topic 3: Partial Fractions Decomposition

Partial fractions is the process of splitting a single, complex rational expression (a fraction with polynomials) back into a sum of simpler fractions.

Why do we need this? This technique is absolutely essential for integrating rational functions in P3 Integration, as simple fractions are much easier to integrate (often leading to logarithmic functions).

Prerequisite Check: Proper vs. Improper Fractions

You can only start the decomposition if the rational function is a proper fraction.

• Proper Fraction: Degree of Numerator < Degree of Denominator. (E.g., \(\frac{x^2}{x^3+1}\))
• Improper Fraction: Degree of Numerator \(\ge\) Degree of Denominator. (E.g., \(\frac{x^3}{x^2+1}\))

If the fraction is improper, you must first perform polynomial long division to express it as:
$$ \text{Polynomial} + \frac{\text{Proper Fraction}}{\text{Denominator}} $$ You then only decompose the proper fraction part.

The Three Standard Forms (P3 Syllabus)

We recall an appropriate form for expressing rational functions based on the factors in the denominator:

Case 1: Distinct Linear Factors

The denominator contains linear factors that are all different.
Form: \( \frac{P(x)}{(ax + b)(cx + d)(ex + f)} \equiv \frac{A}{ax + b} + \frac{B}{cx + d} + \frac{C}{ex + f} \)

Method Tip: Use the Substitution Method (or "Cover-Up" rule) to quickly find A, B, and C by substituting the roots of the denominator (e.g., \( x = -b/a \)) into the derived equation.

Case 2: Repeated Linear Factors

The denominator contains a linear factor raised to a power (e.g., squared).
Form: \( \frac{P(x)}{(ax + b)(cx + d)^2} \equiv \frac{A}{ax + b} + \frac{B}{cx + d} + \frac{C}{(cx + d)^2} \)

Crucial Point: You need a term for every power of the repeated factor, up to the highest power. You cannot combine the B and C terms.

Case 3: Irreducible Quadratic Factors

The denominator contains a quadratic factor that cannot be factorized into real linear factors (e.g., \( x^2 + 4 \)).
Form: \( \frac{P(x)}{(ax + b)(cx^2 + d)} \equiv \frac{A}{ax + b} + \frac{Bx + C}{cx^2 + d} \)

Crucial Point: When the denominator is an irreducible quadratic, the numerator above it must be a linear term (\( Bx + C \)).

Method Tip: For Cases 2 and 3, you usually need a combination of the Substitution Method (for linear factors) and the Equating Coefficients Method (comparing coefficients of \( x^2, x, \) and constant terms) to solve for A, B, and C.

Key Topic 4: Binomial Expansion (Rational Index)

In P1, you learned the Binomial Theorem for positive integer indices \( n \), which resulted in a finite number of terms. In P3, we extend this to when \( n \) is any rational number (negative integers, fractions, etc.).

The Infinite Binomial Series

When \( n \) is rational (but not a positive integer), the expansion results in an infinite series.

Standard Formula (MF19): $$ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots $$

Condition for Validity

Because the series is infinite, it will only converge (be accurate) if the term being raised to the power is small enough.

The expansion is only valid if the modulus of the variable term is less than 1: $$ |x| < 1 \quad \text{or} \quad -1 < x < 1 $$

You must state the range of \( x \) for which the expansion is valid in your answers.

Adapting the Standard Series

The standard formula requires the first term inside the bracket to be 1: \((1 + x)^n\). If you encounter an expression like \((a + x)^n\), you must factor out the first term, \( a \).

Step-by-Step Adaptation:

1. Start with \((a + x)^n\).
2. Factor out \( a \): \( a^n \left(1 + \frac{x}{a}\right)^n \)
3. Let \( Y = \frac{x}{a} \). The expression becomes: \( a^n (1 + Y)^n \)
4. Expand \( (1 + Y)^n \) using the standard formula.
5. Substitute \( Y = \frac{x}{a} \) back into the result.

Example: Expand \( (4 - 3x)^{-1/2} \).
1. Factor out 4: \( (4 - 3x)^{-1/2} = 4^{-1/2} \left(1 - \frac{3x}{4}\right)^{-1/2} \)
2. Simplify \( 4^{-1/2} \): \( 4^{-1/2} = \frac{1}{\sqrt{4}} = \frac{1}{2} \)
3. The expansion is: \( \frac{1}{2} \left(1 + (-\frac{1}{2})(-\frac{3x}{4}) + \dots \right) \)

Validity for Adapted Series

If you use \( Y = \frac{x}{a} \), the validity condition applies to \( Y \):
$$ |Y| < 1 \implies \left|\frac{x}{a}\right| < 1 \implies |x| < |a| $$

In the example above, the validity condition is: \( \left|-\frac{3x}{4}\right| < 1 \implies \frac{3|x|}{4} < 1 \implies |x| < \frac{4}{3} \).

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Key Takeaway for Algebra (P3): This chapter provides the sophisticated tools (Partial Fractions, Generalized Binomial) necessary for the Calculus sections of Paper 3. Practice identifying the correct forms for partial fractions and mastering the factorizing technique required for non-\((1 + x)^n\) binomial expansions.