Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 2 (Paper 2)
Chapter 2.1: Algebra Study Notes
Hello! Welcome to the Algebra chapter for Pure Mathematics 2. This section is all about deepening your understanding of two core concepts: Modulus Functions (or Absolute Value) and Polynomials.
Don't worry if these topics felt abstract before. We'll break them down into practical steps. Mastering this algebra is crucial—it provides the foundational skills needed for success in later topics like Calculus and Trigonometry!
1. Understanding the Modulus Function (Absolute Value)
What is the Modulus?
The modulus of a number, denoted by \(|x|\), represents its distance from zero on the number line.
Because distance can never be negative, the modulus function always gives a positive result (or zero).
Analogy: The Distance Calculator
Imagine your GPS calculates distances. Whether you drive 5 km East (+5) or 5 km West (-5), the distance calculated is always 5 km. The modulus function is the mathematical equivalent of this distance calculator.
Mathematically:
- If \(x \ge 0\), then \(|x| = x\)
- If \(x < 0\), then \(|x| = -x\)
Sketching the Graph of \(y = |ax + b|\)
The graphs of modulus functions are distinctive because of the sharp 'V' shape, caused by the negative part of the original line being reflected upwards.
To sketch \(y = |ax + b|\) (e.g., \(y = |2x - 4|\)):
- Sketch the original linear function: Sketch \(y = ax + b\) as a normal straight line (e.g., \(y = 2x - 4\)).
- Find the root: Determine where the line crosses the x-axis (where \(ax + b = 0\)). This is the vertex (the sharp point) of the V-shape. (In our example, \(2x - 4 = 0\), so \(x = 2\)).
- Reflect the negative part: Any part of the line that lies below the x-axis must be reflected upwards (or reflected in the x-axis). The positive part remains unchanged.
Did you know? The vertex of \(y = |ax + b|\) is always the point \((-\frac{b}{a}, 0)\). This point is crucial as it's where the gradient suddenly changes.
Solving Modulus Equations
When solving equations involving the modulus, we rely on the core identity:
Key Identity: \(\mathbf{|a| = |b| \Leftrightarrow a^2 = b^2}\)
This identity provides the simplest and safest way to solve modulus equations.
Step-by-Step Example: Solve \(|3x - 2| = |2x + 7|\)
- Square both sides: \((3x - 2)^2 = (2x + 7)^2\)
- Expand: \(9x^2 - 12x + 4 = 4x^2 + 28x + 49\)
- Rearrange to a quadratic equation: \(5x^2 - 40x - 45 = 0\)
- Simplify and Factorise (or use the formula): \(x^2 - 8x - 9 = 0 \Rightarrow (x - 9)(x + 1) = 0\)
- Solutions: \(x = 9\) or \(x = -1\)
Alternative Method (Using Cases): You could also consider two cases for the original equation:
- Case 1: \(3x - 2 = 2x + 7\) (Both sides positive or equivalent sign)
- Case 2: \(3x - 2 = -(2x + 7)\) (Opposite signs)
Solving Modulus Inequalities
When solving modulus inequalities, the structure is key. You must use the fact that the modulus relates to distance from a central point.
Type 1: Less Than (\(
If \(|x - a| < b\), this means the distance between \(x\) and \(a\) must be less than \(b\).
This translates into a single compound inequality:
Rule: \(\mathbf{|x - a| < b \Leftrightarrow a - b < x < a + b}\)
Example: Solve \(|x - 5| < 3\).
Here, \(a=5\) and \(b=3\). The solution is \(5 - 3 < x < 5 + 3\), so \(\mathbf{2 < x < 8}\).
Type 2: Greater Than (\(>b\))
If \(|x - a| > b\), this means the distance between \(x\) and \(a\) must be greater than \(b\).
This translates into two separate inequalities (often called 'or' statements):
Rule: \(\mathbf{|x - a| > b \Leftrightarrow x - a > b}\) OR \(\mathbf{x - a < -b}\)
Example: Solve \(|2x + 1| \ge 5\).
Case 1: \(2x + 1 \ge 5 \Rightarrow 2x \ge 4 \Rightarrow \mathbf{x \ge 2}\)
Case 2: \(2x + 1 \le -5 \Rightarrow 2x \le -6 \Rightarrow \mathbf{x \le -3}\)
Quick Review: Modulus Key Takeaways
1. Modulus equations are usually best solved by squaring both sides.
2. Inequalities resulting in 'less than' lead to a single compound range.
3. Inequalities resulting in 'greater than' lead to two separate ranges ('or' statement).
2. Polynomials: Division, Factors, and Remainders
Polynomials are the building blocks of many mathematical functions. In PM2, you need to be comfortable manipulating polynomials up to degree 4.
Polynomial Division
Division is essential for finding factors and roots. You must be able to divide a polynomial \(P(x)\) by either a linear factor (like \(x-a\)) or a quadratic factor (like \(x^2 + 2x + 1\)).
The Division Statement
When you divide a polynomial \(P(x)\) by a divisor \(D(x)\), you get a quotient \(Q(x)\) and a remainder \(R\).
Formula: \(P(x) = D(x) \times Q(x) + R(x)\)
Step-by-Step: Long Division
Example: Divide \(P(x) = x^3 - 5x^2 + 7x - 3\) by \(D(x) = x - 1\).
- Focus on the highest powers: How many times does \(x\) (from \(x-1\)) go into \(x^3\)? It's \(x^2\). Write \(x^2\) in the quotient.
- Multiply: Multiply the divisor by the quotient term: \(x^2(x - 1) = x^3 - x^2\).
- Subtract: Subtract this result from the polynomial. \((x^3 - 5x^2) - (x^3 - x^2) = -4x^2\).
- Bring down the next term: Bring down \(+7x\). You now have \(-4x^2 + 7x\).
- Repeat: How many times does \(x\) go into \(-4x^2\)? It's \(-4x\).
- Continue the process until the remainder's degree is less than the divisor's degree.
Common Mistake: Be extremely careful with signs when subtracting polynomials! It's easy to make a small error here that throws off the entire calculation.
The Remainder Theorem
The Remainder Theorem is a fantastic shortcut! It tells you the remainder of a polynomial division without having to perform the long division.
Theorem 1 (Linear Divisor): When a polynomial \(P(x)\) is divided by a linear factor \((x - a)\), the remainder is \(P(a)\).
Example: Find the remainder when \(P(x) = x^3 + x - 1\) is divided by \((x - 2)\).
We substitute \(x = 2\): \(P(2) = (2)^3 + 2 - 1 = 8 + 2 - 1 = 9\). The remainder is 9.
Theorem 2 (General Linear Divisor): When \(P(x)\) is divided by a linear factor \((ax + b)\), the remainder is \(P(-\frac{b}{a})\).
This second theorem is vital for Paper 2, as questions often involve divisors where the coefficient of \(x\) is not 1.
Example: Find the remainder when \(P(x) = 2x^2 + 3x + 1\) is divided by \((2x - 1)\).
Set the divisor to zero: \(2x - 1 = 0 \Rightarrow x = 1/2\).
Remainder is \(P(1/2) = 2(1/2)^2 + 3(1/2) + 1 = 2(1/4) + 3/2 + 1 = 1/2 + 3/2 + 1 = 2 + 1 = \mathbf{3}\).
The Factor Theorem
The Factor Theorem is simply a special case of the Remainder Theorem.
Theorem: \((x - a)\) is a factor of \(P(x)\) if and only if \(P(a) = 0\).
In other words, if substituting a value \(a\) into the polynomial makes the result zero, then \(a\) is a root, and \((x - a)\) is a factor.
We use the Factor Theorem to:
- Find unknown coefficients in a polynomial (e.g., finding \(k\) if \((x-3)\) is a known factor).
- Solve polynomial equations (e.g., finding the first factor by inspection, then using division to find the remaining quadratic factor).
Tip for Finding the First Factor:
If a polynomial \(P(x)\) has integer coefficients, the integer factors are almost always factors of the constant term.
If \(P(x) = x^3 + 2x^2 - 5x - 6\), try substituting \(x = \pm 1, \pm 2, \pm 3, \pm 6\).
\(P(2) = 8 + 8 - 10 - 6 = 0\). Since \(P(2) = 0\), \((x - 2)\) is a factor!
Once you find one linear factor, you use polynomial division to reduce the polynomial to a simpler quadratic expression, which you can then solve easily using standard methods (factorising or formula).
Key Takeaway: Polynomials
The Remainder and Factor Theorems are essential tools. Use the Factor Theorem to identify roots (where \(R=0\)) and Polynomial Division to find the quotient and complete the factorisation. Always remember the form \(P(-\frac{b}{a})\) when the factor is \((ax+b)\).
If \(|x - a| < b\), this means the distance between \(x\) and \(a\) must be less than \(b\).
This translates into a single compound inequality:
Rule: \(\mathbf{|x - a| < b \Leftrightarrow a - b < x < a + b}\)
Example: Solve \(|x - 5| < 3\).
Here, \(a=5\) and \(b=3\). The solution is \(5 - 3 < x < 5 + 3\), so \(\mathbf{2 < x < 8}\).
Type 2: Greater Than (\(>b\))
If \(|x - a| > b\), this means the distance between \(x\) and \(a\) must be greater than \(b\).
This translates into two separate inequalities (often called 'or' statements):
Rule: \(\mathbf{|x - a| > b \Leftrightarrow x - a > b}\) OR \(\mathbf{x - a < -b}\)
Example: Solve \(|2x + 1| \ge 5\).
Case 1: \(2x + 1 \ge 5 \Rightarrow 2x \ge 4 \Rightarrow \mathbf{x \ge 2}\)
Case 2: \(2x + 1 \le -5 \Rightarrow 2x \le -6 \Rightarrow \mathbf{x \le -3}\)
Quick Review: Modulus Key Takeaways
1. Modulus equations are usually best solved by squaring both sides.
2. Inequalities resulting in 'less than' lead to a single compound range.
3. Inequalities resulting in 'greater than' lead to two separate ranges ('or' statement).
2. Polynomials: Division, Factors, and Remainders
Polynomials are the building blocks of many mathematical functions. In PM2, you need to be comfortable manipulating polynomials up to degree 4.
Polynomial Division
Division is essential for finding factors and roots. You must be able to divide a polynomial \(P(x)\) by either a linear factor (like \(x-a\)) or a quadratic factor (like \(x^2 + 2x + 1\)).
The Division Statement
When you divide a polynomial \(P(x)\) by a divisor \(D(x)\), you get a quotient \(Q(x)\) and a remainder \(R\).
Formula: \(P(x) = D(x) \times Q(x) + R(x)\)
Step-by-Step: Long Division
Example: Divide \(P(x) = x^3 - 5x^2 + 7x - 3\) by \(D(x) = x - 1\).
- Focus on the highest powers: How many times does \(x\) (from \(x-1\)) go into \(x^3\)? It's \(x^2\). Write \(x^2\) in the quotient.
- Multiply: Multiply the divisor by the quotient term: \(x^2(x - 1) = x^3 - x^2\).
- Subtract: Subtract this result from the polynomial. \((x^3 - 5x^2) - (x^3 - x^2) = -4x^2\).
- Bring down the next term: Bring down \(+7x\). You now have \(-4x^2 + 7x\).
- Repeat: How many times does \(x\) go into \(-4x^2\)? It's \(-4x\).
- Continue the process until the remainder's degree is less than the divisor's degree.
Common Mistake: Be extremely careful with signs when subtracting polynomials! It's easy to make a small error here that throws off the entire calculation.
The Remainder Theorem
The Remainder Theorem is a fantastic shortcut! It tells you the remainder of a polynomial division without having to perform the long division.
Theorem 1 (Linear Divisor): When a polynomial \(P(x)\) is divided by a linear factor \((x - a)\), the remainder is \(P(a)\).
Example: Find the remainder when \(P(x) = x^3 + x - 1\) is divided by \((x - 2)\).
We substitute \(x = 2\): \(P(2) = (2)^3 + 2 - 1 = 8 + 2 - 1 = 9\). The remainder is 9.
Theorem 2 (General Linear Divisor): When \(P(x)\) is divided by a linear factor \((ax + b)\), the remainder is \(P(-\frac{b}{a})\).
This second theorem is vital for Paper 2, as questions often involve divisors where the coefficient of \(x\) is not 1.
Example: Find the remainder when \(P(x) = 2x^2 + 3x + 1\) is divided by \((2x - 1)\).
Set the divisor to zero: \(2x - 1 = 0 \Rightarrow x = 1/2\).
Remainder is \(P(1/2) = 2(1/2)^2 + 3(1/2) + 1 = 2(1/4) + 3/2 + 1 = 1/2 + 3/2 + 1 = 2 + 1 = \mathbf{3}\).
The Factor Theorem
The Factor Theorem is simply a special case of the Remainder Theorem.
Theorem: \((x - a)\) is a factor of \(P(x)\) if and only if \(P(a) = 0\).
In other words, if substituting a value \(a\) into the polynomial makes the result zero, then \(a\) is a root, and \((x - a)\) is a factor.
We use the Factor Theorem to:
- Find unknown coefficients in a polynomial (e.g., finding \(k\) if \((x-3)\) is a known factor).
- Solve polynomial equations (e.g., finding the first factor by inspection, then using division to find the remaining quadratic factor).
Tip for Finding the First Factor:
If a polynomial \(P(x)\) has integer coefficients, the integer factors are almost always factors of the constant term.
If \(P(x) = x^3 + 2x^2 - 5x - 6\), try substituting \(x = \pm 1, \pm 2, \pm 3, \pm 6\).
\(P(2) = 8 + 8 - 10 - 6 = 0\). Since \(P(2) = 0\), \((x - 2)\) is a factor!
Once you find one linear factor, you use polynomial division to reduce the polynomial to a simpler quadratic expression, which you can then solve easily using standard methods (factorising or formula).
Key Takeaway: Polynomials
The Remainder and Factor Theorems are essential tools. Use the Factor Theorem to identify roots (where \(R=0\)) and Polynomial Division to find the quotient and complete the factorisation. Always remember the form \(P(-\frac{b}{a})\) when the factor is \((ax+b)\).