CIE AS/A Level Maths-2.1 Algebra- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-2.1 Algebra- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-2.1 Algebra- Study Notes
Key Concepts:
- Absolute Value (Modulus) Functions
- Polynomial Division
- Factor Theorem and Remainder Theorem
Absolute Value (Modulus) Functions
Absolute Value (Modulus) Functions
Meaning of |x|
The modulus (absolute value) of a real number \(x\), written \(|x|\), is its distance from zero on the number line. Thus:
- \(|x| = x\) if \(x \geq 0\)
- \(|x| = -x\) if \(x < 0\)
Example: \(|5| = 5\), \(|-7| = 7\).
Graph of \(y = |ax+b|\)
The graph of \(y=|ax+b|\) can be understood as a transformation of the straight line \(y=ax+b\):
- Where \(ax+b \geq 0\), the graph is the same as the straight line.
- Where \(ax+b < 0\), the graph is reflected above the x-axis (values become positive).
This produces a “V-shaped” graph, with the corner (vertex) at the point where \(ax+b=0\).
. Key modulus relations
- \(|a| = |b| \iff a^2 = b^2\)
Because both sides are non-negative and squaring removes the modulus.
- \(|x-a| < b \iff a-b < x < a+b\)
This expresses the idea that \(x\) lies within a distance \(b\) of \(a\).
- \(|x-a| > b \iff x < a-b \; \text{or} \; x > a+b\)
This means \(x\) lies further than distance \(b\) from \(a\).
Solving equations and inequalities with modulus
General strategies:
- Case splitting: Replace \(|f(x)|\) with \(f(x)\) if \(f(x)\geq 0\), or \(-f(x)\) if \(f(x)<0\).
- Using standard relations: Apply results like \(|a|=|b|\iff a^2=b^2\) or \(|x-a|<b\iff a-b<x<a+b\).
- Graphical method: For simple inequalities, sketch the modulus graph and read solutions directly.
Example:
Solve the equation \(|2x – 3| = 5\).
▶️ Answer/Explanation
Step 1: Use the property
\(|a| = b \iff a = b \;\; \text{or} \;\; a = -b\).
Step 2: Apply to the given equation
\(2x – 3 = 5 \quad \text{or} \quad 2x – 3 = -5\).
Step 3: Solve each case
Case 1: \(2x – 3 = 5 \implies 2x = 8 \implies x = 4\).
Case 2: \(2x – 3 = -5 \implies 2x = -2 \implies x = -1\).
Final Answer: \(x = -1\) or \(x = 4\).
Example:
Solve the inequality \(|x-2| < 3\).
▶️ Answer/Explanation
Step 1: Recall the property
\(|x-a| < b \iff a-b < x < a+b\).
Step 2: Apply with \(a=2\), \(b=3\)
\(2 – 3 < x < 2 + 3\).
\(-1 < x < 5\).
Final Answer: The solution set is \(-1 < x < 5\).
Example:
Sketch the graph of \(y = |x+1|\).
▶️ Answer/Explanation
Step 1: Recall definition
\(|f(x)| = f(x)\) if \(f(x)\geq 0\), and \(|f(x)|=-f(x)\) if \(f(x)<0\).
Step 2: Find when inside is zero
\(x+1 = 0 \implies x = -1\).
Step 3: Piecewise definition
If \(x \geq -1\), \(y = x+1\).
If \(x < -1\), \(y = -(x+1) = -x-1\).
Step 4: Sketch
The graph is a “V” with vertex at \((-1,0)\), slope \(1\) to the right, slope \(-1\) to the left.
Example:
A point \(x\) on the real line satisfies \(|x-5| > 2\). Describe the possible positions of \(x\).
▶️ Answer/Explanation
Step 1: Recall property
\(|x-a| > b \iff x < a-b \; \text{or} \; x > a+b\).
Step 2: Apply with \(a=5\), \(b=2\)
\(x < 3 \;\; \text{or} \;\; x > 7\).
Step 3: Interpret
The point \(x\) lies more than 2 units away from 5 on the number line.
Final Answer: \(x \in (-\infty,3) \cup (7,\infty)\).
Polynomial Division
Polynomial Division
Polynomial division is a process similar to long division of numbers, but applied to algebraic expressions. It allows us to divide one polynomial by another (usually linear or quadratic), and express the result in the form:
$ f(x) = d(x) \cdot q(x) + r(x) $
where:
- \(f(x)\) is the dividend (the polynomial being divided).
- \(d(x)\) is the divisor (the polynomial dividing).
- \(q(x)\) is the quotient (result of division).
- \(r(x)\) is the remainder (must have degree smaller than the divisor).
Key Points:
- If divisor is linear (\(ax+b\)), remainder is a constant.
- If divisor is quadratic (\(ax^2+bx+c\)), remainder is linear.
- The remainder theorem states: if dividing by \(x-a\), the remainder is \(f(a)\).
Example:
Divide \(f(x) = 2x^3 + 3x^2 – 5x + 6\) by \((x-2)\). Identify the quotient and remainder.
▶️ Answer/Explanation
Step 1: Apply synthetic division (since divisor is linear)
Dividing by \(x-2\) means \(a=2\).
Coefficients: \(2, 3, -5, 6\).
Synthetic division process:
Step 2: Interpret
Quotient = \(2x^2 + 7x + 9\).
Remainder = 24.
Final Answer:
\(2x^3 + 3x^2 – 5x + 6 = (x-2)(2x^2 + 7x + 9) + 24\).
Example:
Divide \(f(x) = x^4 + 2x^3 – 3x^2 + 4x – 5\) by \(x^2+1\). Identify the quotient and remainder.
▶️ Answer/Explanation
Step 1: Perform polynomial long division
Divide \(x^4 + 2x^3 – 3x^2 + 4x – 5\) by \(x^2+1\).
\(x^4 ÷ x^2 = x^2\). Multiply divisor: \(x^2(x^2+1) = x^4 + x^2\).
Subtract: \((x^4 + 2x^3 – 3x^2) – (x^4 + x^2) = 2x^3 – 4x^2\).
Bring down terms → \(2x^3 – 4x^2 + 4x – 5\).
\(2x^3 ÷ x^2 = 2x\). Multiply divisor: \(2x(x^2+1) = 2x^3 + 2x\).
Subtract: \((2x^3 – 4x^2 + 4x) – (2x^3 + 2x) = -4x^2 + 2x\).
Bring down → \(-4x^2 + 2x – 5\).
\(-4x^2 ÷ x^2 = -4\). Multiply divisor: \(-4(x^2+1) = -4x^2 -4\).
Subtract: \((-4x^2 + 2x – 5) – (-4x^2 -4) = 2x -1\).
Step 2: Interpret
Quotient = \(x^2 + 2x – 4\).
Remainder = \(2x – 1\) (degree less than divisor).
Final Answer:
\(x^4 + 2x^3 – 3x^2 + 4x – 5 = (x^2+1)(x^2 + 2x – 4) + (2x -1)\).
Factor Theorem and Remainder Theorem
Factor Theorem and Remainder Theorem
Polynomial division, factorization, and evaluation are closely connected through two important results:
Remainder Theorem
When a polynomial \(f(x)\) is divided by a linear divisor \((x-a)\), the remainder is simply \(f(a)\).
- If \(f(a) = r\), then remainder is \(r\).
- If \(f(a) = 0\), then \((x-a)\) is a factor of \(f(x)\).
Example idea: If dividing \(f(x)\) by \((x-3)\), just evaluate \(f(3)\) to get the remainder.
Factor Theorem
The factor theorem is a special case of the remainder theorem. It states:
If \(f(a) = 0\), then \((x-a)\) is a factor of \(f(x)\).
This theorem allows us to find factors of polynomials and solve polynomial equations.
Extension to factors of the form \((ax+b)\)
If the divisor is of the form \((ax+b)\), then set it equal to zero:
\(ax+b=0 \implies x=-\dfrac{b}{a}\).
Then evaluate \(f\!\left(-\dfrac{b}{a}\right)\) to find the remainder. If it equals 0, then \((ax+b)\) is a factor.
Using the theorems to find unknown coefficients
If a polynomial contains unknown coefficients, substituting values of \(x\) using the factor or remainder theorem creates equations to solve for the unknowns.
Example:
Find the remainder when \(f(x) = 2x^3 – 5x + 4\) is divided by \((x-2)\).
▶️ Answer/Explanation
By the remainder theorem, remainder = \(f(2)\).
\(f(2) = 2(2^3) – 5(2) + 4 = 16 – 10 + 4 = 10\).
Final Answer: Remainder = 10.
Example:
Show that \((x-3)\) is a factor of \(f(x) = x^3 – 7x + 6\), and hence factorize the polynomial completely.
▶️ Answer/Explanation
Step 1: Apply factor theorem
\(f(3) = 3^3 – 7(3) + 6 = 27 – 21 + 6 = 12\).
\(f(3) \neq 0\), so wait, check carefully:
\(f(x) = x^3 – 7x + 6\).
\(f(3) = 27 – 21 + 6 = 12\) → correction: (x-3) is not a factor.
Try instead \((x-1)\): \(f(1) = 1 – 7 + 6 = 0\).
So \((x-1)\) is a factor.
Step 2: Divide polynomial
Divide \(x^3 – 7x + 6\) by \((x-1)\) → quotient = \(x^2 + x – 6\).
Factor further: \(x^2 + x – 6 = (x+3)(x-2)\).
Final Answer:
\(f(x) = (x-1)(x+3)(x-2)\).
Example:
Find the remainder when \(f(x) = 3x^3 + 2x^2 – x + 5\) is divided by \((2x+1)\).
▶️ Answer/Explanation
Step 1: Convert divisor
\(2x+1=0 \implies x=-\dfrac{1}{2}\).
Step 2: Apply remainder theorem
Remainder = \(f\!\left(-\dfrac{1}{2}\right)\).
\(f(-\tfrac{1}{2}) = 3\left(-\dfrac{1}{2}\right)^3 + 2\left(-\dfrac{1}{2}\right)^2 – \left(-\dfrac{1}{2}\right) + 5\).
= \(3\left(-\dfrac{1}{8}\right) + 2\left(\dfrac{1}{4}\right) + \dfrac{1}{2} + 5\).
= \(-\dfrac{3}{8} + \dfrac{1}{2} + \dfrac{1}{2} + 5\).
= \(-\dfrac{3}{8} + 6\).
= \(\dfrac{45}{8}\).
Final Answer: Remainder = \(\dfrac{45}{8}\).
Example:
The polynomial \(f(x) = x^3 + ax^2 + bx + 6\) has a factor \((x-2)\). Find \(a\) and \(b\).
▶️ Answer/Explanation
Step 1: Apply factor theorem
If \((x-2)\) is a factor, then \(f(2)=0\).
\(f(2) = (2)^3 + a(2^2) + b(2) + 6 = 8 + 4a + 2b + 6 = 14 + 4a + 2b\).
So: \(14 + 4a + 2b = 0 \implies 2a + b = -7\). (Equation 1)
Step 2: Additional condition
Suppose also \((x+1)\) is a factor (if stated in question). Then \(f(-1)=0\).
\(f(-1) = (-1)^3 + a(-1)^2 + b(-1) + 6 = -1 + a – b + 6 = a – b + 5\).
So: \(a – b + 5 = 0 \implies a – b = -5\). (Equation 2)
Step 3: Solve
From (1) and (2):
2a + b = -7
a – b = -5
Add equations: 3a = -12 → a = -4.
Substitute: -4 – b = -5 → b = 1.
Final Answer: \(a=-4\), \(b=1\).