Welcome to Pure Mathematics 3: Algebra and Functions!

Hello! You've arrived at Unit P3, and this chapter on Algebra and Functions is truly a powerhouse. While some concepts might feel familiar from P1 or P2, we are expanding into more complex, essential tools—particularly Partial Fractions and the Modulus Function.

Don't worry if this seems tricky at first! We will break down every concept into clear, actionable steps. Mastering this material is crucial because Partial Fractions are often the key needed to successfully integrate complex functions later in P3, and the Modulus Function introduces a new type of graphing and equation-solving skill set. Let's get started!

Section 1: Algebraic Fractions and Partial Fractions

1.1 Proper vs. Improper Fractions (Quick Review)

Before we dive into splitting fractions, we must make sure our algebraic fraction is a proper fraction.

  • Proper Fraction: The highest power (degree) of \(x\) in the numerator is strictly less than the highest power of \(x\) in the denominator. Example: \(\frac{2x + 1}{x^2 + 5x - 6}\) (Degree 1 over Degree 2).
  • Improper Fraction: The highest power (degree) of \(x\) in the numerator is greater than or equal to the highest power of \(x\) in the denominator. Example: \(\frac{x^3 - 2x + 1}{x^2 - 1}\) (Degree 3 over Degree 2).

Crucial Step: If you have an improper fraction, you must use algebraic division first.

If \(\frac{A(x)}{B(x)}\) is improper, you must divide it so it looks like:

\[ \frac{A(x)}{B(x)} = Q(x) + \frac{R(x)}{B(x)} \]

where \(Q(x)\) is the quotient (the polynomial part) and \(\frac{R(x)}{B(x)}\) is now the proper fraction that you can split into partial fractions.

1.2 Introduction to Partial Fractions

Partial fractions are the process of reversing the addition of fractions. We take a single, complicated fraction and split it into two or more simpler fractions that are easier to work with (especially for integration).

Case A: Distinct Linear Factors in the Denominator

This is the most common case. If the denominator factors into unique linear terms, \((ax+b)\) and \((cx+d)\), the structure of the partial fractions is straightforward.

Setup Rule: For a fraction \(\frac{P(x)}{(x+a)(x-b)}\), we write:

\[ \frac{P(x)}{(x+a)(x-b)} \equiv \frac{A}{x+a} + \frac{B}{x-b} \]

Step-by-Step Method:

  1. Set up the identity (as above).
  2. Multiply the entire identity by the original denominator, \((x+a)(x-b)\). This clears the denominators.
  3. Method 1: Substitution (The Easy Way)

    Substitute values of \(x\) that make the brackets zero (e.g., \(x = -a\) and \(x = b\)). This cancels out one unknown variable (\(A\) or \(B\)) immediately, allowing you to solve for the other.

  4. Method 2: Equating Coefficients

    Expand the right side and collect terms for \(x^2\), \(x\), and constants. Set the coefficients of these terms equal to the corresponding coefficients in the original numerator. This gives you a system of simultaneous equations.

Tip for Speed: Always use the substitution method first to find as many constants as possible! Only resort to equating coefficients if substitution runs out of factors (e.g., in Case B or C).

Case B: Repeated Linear Factors in the Denominator

If a factor appears twice, like \((x+a)^2\), we must account for both the single factor and the squared factor. This is a common place for students to make a mistake!

Setup Rule: For a fraction \(\frac{P(x)}{(x+a)^2(x-b)}\), we write:

\[ \frac{P(x)}{(x+a)^2(x-b)} \equiv \frac{A}{x+a} + \frac{B}{(x+a)^2} + \frac{C}{x-b} \]

Example: If the denominator is \((x-1)^3\), you would need three terms: \(\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}\).

Solving Strategy:

  1. Use substitution to find \(B\) and \(C\). (Set \(x=1\) to find \(C\), set \(x=b\) to find \(B\)).
  2. Once substitution is complete, you must use equating coefficients (e.g., equate the coefficient of the highest power of \(x\)) or substitute a random easy number (like \(x=0\)) to find the final unknown constant \(A\).
Case C: Irreducible Quadratic Factors in the Denominator

An irreducible quadratic factor is a quadratic expression (like \(x^2 + 4\)) that cannot be factored further using real numbers (i.e., it has no real roots).

Setup Rule: For a fraction \(\frac{P(x)}{(x^2+ax+b)(x-c)}\), where \(x^2+ax+b\) is irreducible, we write:

\[ \frac{P(x)}{(x^2+ax+b)(x-c)} \equiv \frac{Ax+B}{x^2+ax+b} + \frac{C}{x-c} \]

Notice that the numerator over the quadratic factor must be linear (\(Ax+B\)).

Solving Strategy:

  1. Use substitution on the linear factor to find \(C\). (e.g., set \(x=c\)).
  2. Use equating coefficients (especially the highest power, like \(x^3\), and the constant term) to set up simultaneous equations to find \(A\) and \(B\).
Key Takeaway: Partial Fractions

Always check if the fraction is proper first. The structure of your partial fraction identity must match the factors in the denominator. A linear factor gets a constant numerator; a quadratic factor gets a linear numerator (\(Ax+B\)).

Section 2: The Modulus Function (Absolute Value)

The modulus function, denoted by \(|x|\), gives the absolute value of \(x\). Conceptually, it represents the distance of a number from zero, and distance is always positive!

2.1 Definition and Notation

The modulus function is defined piecewise:

\[ |x| = \begin{cases}\nx & \text{if } x \ge 0 \ -x & \text{if } x < 0 \end{cases} \]

Example: \(|5| = 5\). But \(|-5| = -(-5) = 5\).

2.2 Graphs involving the Modulus Function

Understanding the geometry of modulus graphs is vital, both for sketching and for solving equations graphically.

Transformation 1: \(y = |f(x)|\)

This means you apply the modulus after calculating the function's value.

Rule: The graph of \(y = |f(x)|\) is obtained by reflecting any part of \(y = f(x)\) that lies below the x-axis into the positive region (above the x-axis). The negative parts are flipped up.

Analogy: Imagine the x-axis is a hinge. Everything below the hinge flips up.

Transformation 2: \(y = f(|x|)\)

This means you apply the modulus to the input variable, \(x\).

Rule: The graph of \(y = f(|x|)\) is always symmetrical about the y-axis.

  1. Sketch the graph for \(x \ge 0\).
  2. Erase the graph for \(x < 0\).
  3. Reflect the graph from \(x \ge 0\) across the y-axis to fill the \(x < 0\) region.

Did you know? Functions where \(f(x) = f(-x)\) are called even functions, and they always exhibit y-axis symmetry, which is exactly what \(y = f(|x|)\) creates!

2.3 Solving Equations involving Modulus

When solving equations like \(|ax + b| = c\) or \(|f(x)| = |g(x)|\), you have two reliable methods.

Method 1: Splitting into Cases (Recommended for Simplicity)

Since the expression inside the modulus bars could be positive or negative, we test both possibilities.

Step-by-Step (for \(|ax + b| = cx + d\)):

  1. Case 1 (Positive): Assume the expression inside is positive or zero. \[ ax + b = cx + d \]
  2. Case 2 (Negative): Assume the expression inside is negative. We must place a negative sign in front of the expression. \[ -(ax + b) = cx + d \]
  3. Solve both resulting linear equations to find all potential solutions for \(x\).
  4. Check Solutions: Always substitute your answers back into the original equation, especially if the right-hand side also contains \(x\). Modulus problems can sometimes produce extraneous solutions (solutions that satisfy the derived equations but not the original one).
Method 2: Squaring Both Sides

If you have modulus signs on both sides, squaring is often efficient because \(|A|^2 = A^2\).

Example: \(|x + 1| = |2x - 5|\)
\[ (x + 1)^2 = (2x - 5)^2 \]

This simplifies the equation into a standard quadratic, which you solve normally. This method always works for equations and avoids extraneous solutions when both sides are already positive (i.e., both are modulus functions).

2.4 Solving Inequalities involving Modulus

Solving modulus inequalities requires even more care than solving equations. We rely heavily on the definition and understanding the resulting graph.

Inequalities with a Constant (A quick trick)

If \(a\) is a positive constant:

  • "Less than" (AND/Inside): \(|x| < a\) means \(x\) is within \(a\) distance of zero. \[ |x| < a \implies -a < x < a \]
  • "Greater than" (OR/Outside): \(|x| > a\) means \(x\) is further than \(a\) distance from zero. \[ |x| > a \implies x > a \quad \text{OR} \quad x < -a \]
Inequalities with \(x\) on Both Sides (The only safe method)

When solving inequalities like \(|2x - 1| \le x + 4\), the squaring method is generally preferred, but you must be extremely careful when dealing with the potential signs of the expressions.

Recommended Method: Squaring (But remember the sign restriction!)

For \(|f(x)| \le g(x)\):

  1. Establish the Right-Hand Restriction: Since \(|f(x)|\) is always non-negative, the right-hand side \(g(x)\) must also be non-negative for a solution to exist.

    Requirement: \(g(x) \ge 0\). This constraint must be included in your final solution set.

  2. Square Both Sides: Square both sides and solve the resulting quadratic inequality. \[ (f(x))^2 \le (g(x))^2 \]
  3. Combine Solutions: Your final answer must satisfy both the solutions found in Step 2 AND the restriction found in Step 1.

Alternative Method: Sketching the graphs \(y = |f(x)|\) and \(y = g(x)\) and determining where one graph lies above the other can be the clearest way to solve complex modulus inequalities.

Common Mistake Alert!

When squaring inequalities like \(|2x| < x + 3\), students often forget the necessary prerequisite condition: \(x + 3\) must be positive (\(x + 3 > 0\)). If you forget this constraint, you will gain incorrect solutions.

Key Takeaway: The Modulus Function

Modulus means two cases: positive and negative. For equations, use splitting cases or squaring. For inequalities involving \(x\) on the RHS, use squaring but always enforce the prerequisite constraint that the RHS must be positive/non-negative.