CIE AS/A Level Maths-3.1 Algebra- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-3.1 Algebra- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-3.1 Algebra- Study Notes
Key Concepts:
- Absolute Value Functions and Equations
- Polynomial Division
- Factor Theorem and Remainder Theorem
- Partial Fractions
- Binomial Series
Absolute Value Functions and Equations
Absolute Value Functions and Equations
Definition of Absolute Value
The absolute value of a number \(x\), written as \(|x|\), represents its distance from zero on the number line.
- \(|x| \geq 0\) for all real numbers \(x\).
- \(|x| = x\) if \(x \geq 0\), and \(|x| = -x\) if \(x < 0\).
Graph of \(y = |ax + b|\)
- The graph is V-shaped.
- It is symmetric about the vertical line through its vertex.
- The vertex occurs at \(x = -\dfrac{b}{a}\).
- The slope of the right-hand side is \(+a\), and of the left-hand side is \(-a\).
Important Relations
- \(|a| = |b| \;\;\Leftrightarrow\;\; a^2 = b^2\).
- \(|x-a| < b \;\;\Leftrightarrow\;\; a-b < x < a+b\).
- \(|x-a| > b \;\;\Leftrightarrow\;\; x < a-b \;\text{ or }\; x > a+b\).
Example:
Solve \(|3x – 2| = |2x + 7|\).
▶️ Answer/Explanation
We use the property: \(|A| = |B| \;\;\Rightarrow\;\; A = B \;\text{ or }\; A = -B\).
Case 1: \(3x – 2 = 2x + 7\)
\(x = 9\).
Case 2: \(3x – 2 = -(2x + 7)\)
\(3x – 2 = -2x – 7 \;\;\Rightarrow\;\; 5x = -5 \;\;\Rightarrow\;\; x = -1\).
Final Answer: \(\boxed{x = -1 \text{ or } x = 9}\).
Example:
Solve the inequality \(2x + 5 < |x + 1|\).
▶️ Answer/Explanation
We solve by considering the two cases for \(|x+1|\).
Case 1: If \(x+1 \geq 0 \;\;(x \geq -1)\), then \(|x+1| = x+1\).
Inequality: \(2x + 5 < x+1 \;\;\Rightarrow\;\; x < -4\).
But this contradicts \(x \geq -1\). Hence, no solution in this case.
Case 2: If \(x+1 < 0 \;\;(x < -1)\), then \(|x+1| = -(x+1) = -x-1\).
Inequality: \(2x + 5 < -x – 1\).
\(\;\; 3x < -6 \;\;\Rightarrow\;\; x < -2\).
This is consistent with \(x < -1\).
Final Answer: \(\boxed{x < -2}\).
Polynomial Division
Polynomial Division
Polynomial division works similarly to long division of numbers. We divide one polynomial (the dividend) by another (the divisor), obtaining a quotient and possibly a remainder.
Key Points:
- If a polynomial \(f(x)\) of degree ≤ 4 is divided by a linear or quadratic polynomial, we can find the quotient and remainder.
- The remainder must always have degree < degree of divisor.
- If divisor is linear \((x – a)\), remainder is a constant.
- If divisor is quadratic, remainder is at most linear.
Example:
Divide \(f(x) = 2x^3 + 3x^2 – 5x + 6\) by \((x-2)\). Identify the quotient and remainder.
▶️ Answer/Explanation
Step 1: Perform long or synthetic division.
\((2x^3 + 3x^2 – 5x + 6) \div (x – 2)\)
Quotient: \(2x^2 + 7x + 9\)
Remainder: \(24\)
Final Result:
\(\boxed{2x^3 + 3x^2 – 5x + 6 = (x-2)(2x^2 + 7x + 9) + 24}\)
Example:
Divide \(f(x) = x^4 – 3x^2 + 2x + 5\) by \((x^2 – 1)\).
▶️ Answer/Explanation
Step 1: Perform polynomial long division.
\((x^4 – 3x^2 + 2x + 5) \div (x^2 – 1)\)
Quotient: \(x^2 – 2\)
Remainder: \(2x + 3\)
Final Result:
\(\boxed{x^4 – 3x^2 + 2x + 5 = (x^2 – 1)(x^2 – 2) + (2x + 3)}\)
Factor Theorem and Remainder Theorem
Factor Theorem and Remainder Theorem
Remainder Theorem:
- If a polynomial \(f(x)\) is divided by \((x – a)\), the remainder is \(f(a)\).
- This means we don’t need full long division – we can simply substitute \(x = a\) into the polynomial.
Factor Theorem:
- If \(f(a) = 0\), then \((x – a)\) is a factor of \(f(x)\).
- Conversely, if \((x – a)\) is a factor, then \(f(a) = 0\).
- This theorem is useful for solving polynomial equations and finding unknown coefficients.
Extension:
- For divisors of the form \((ax + b)\), remainder is found by evaluating \(f\!\left(-\dfrac{b}{a}\right)\).
- If \(f\!\left(-\dfrac{b}{a}\right) = 0\), then \((ax + b)\) is a factor.
Example
Find the remainder when \(f(x) = 2x^3 – 5x^2 + 4x – 7\) is divided by \((x – 3)\).
▶️ Answer/Explanation
By the remainder theorem, remainder = \(f(3)\).
\(f(3) = 2(3)^3 – 5(3)^2 + 4(3) – 7\)
\(= 54 – 45 + 12 – 7 = 14\)
Final Answer:
\(\boxed{14}\)
Example
Show that \((x + 2)\) is a factor of \(f(x) = x^3 + 3x^2 – 4\).
▶️ Answer/Explanation
Here, divisor is \((x + 2)\). Put \(x = -2\).
\(f(-2) = (-2)^3 + 3(-2)^2 – 4\)
\(= -8 + 12 – 4 = 0\)
Since \(f(-2) = 0\), by Factor Theorem, \((x + 2)\) is a factor.
Final Answer:
\(\boxed{(x+2)\ \text{is a factor}}\)
Example
The polynomial \(f(x) = x^3 + kx^2 – 2x – 3\) has \((x – 1)\) as a factor. Find \(k\).
▶️ Answer/Explanation
Since \((x – 1)\) is a factor, \(f(1) = 0\).
\(f(1) = (1)^3 + k(1)^2 – 2(1) – 3 = 0\)
\(1 + k – 2 – 3 = 0\)
\(k – 4 = 0 \quad \Rightarrow \quad k = 4\)
Final Answer:
\(\boxed{k = 4}\)
Partial Fractions
Partial Fractions
A rational function can often be expressed as a sum of simpler fractions. This process is called partial fraction decomposition.
We only apply it when the degree of the numerator is less than the degree of the denominator.
Forms of Partial Fractions
- For distinct linear factors:
\(\dfrac{P(x)}{(ax+b)(cx+d)(ex+f)} = \dfrac{A}{ax+b} + \dfrac{B}{cx+d} + \dfrac{C}{ex+f}\)
- For a repeated linear factor:
\(\dfrac{P(x)}{(ax+b)(cx+d)^2} = \dfrac{A}{ax+b} + \dfrac{B}{cx+d} + \dfrac{C}{(cx+d)^2}\)
- For an irreducible quadratic factor:
\(\dfrac{P(x)}{(ax+b)(cx^2+d)} = \dfrac{A}{ax+b} + \dfrac{Bx+C}{cx^2+d}\)
Example:
Decompose: \(\dfrac{2x+3}{(x+1)(x+2)}\).
▶️ Answer/Explanation
Assume: \(\dfrac{2x+3}{(x+1)(x+2)} = \dfrac{A}{x+1} + \dfrac{B}{x+2}\).
Multiply through: \(2x+3 = A(x+2) + B(x+1)\).
Expand: \(2x+3 = (A+B)x + (2A+B)\).
Compare coefficients: \(A+B = 2\), \(2A+B = 3\).
Solving: \(A=1, B=1\).
Final result: \(\dfrac{1}{x+1} + \dfrac{1}{x+2}\).
Example:
Decompose: \(\dfrac{3x+5}{(x+1)(x+2)^2}\).
▶️ Answer/Explanation
Assume: \(\dfrac{3x+5}{(x+1)(x+2)^2} = \dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2}\).
Multiply through: \(3x+5 = A(x+2)^2 + B(x+1)(x+2) + C(x+1)\).
Expand and simplify terms, then compare coefficients of \(x^2, x, \text{ and constant}\).
After solving, we obtain specific values for \(A,B,C\).
Final result: \(\dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2}\).
Example:
Decompose: \(\dfrac{2x+1}{(x+1)(x^2+1)}\).
▶️ Answer/Explanation
Assume: \(\dfrac{2x+1}{(x+1)(x^2+1)} = \dfrac{A}{x+1} + \dfrac{Bx+C}{x^2+1}\).
Multiply through: \(2x+1 = A(x^2+1) + (Bx+C)(x+1)\).
Expand: \(2x+1 = A x^2 + A + Bx^2 + (B+C)x + C\).
Collect: \((A+B)x^2 + (B+C)x + (A+C)\).
Compare coefficients: \(A+B = 0\), \(B+C = 2\), \(A+C = 1\).
Solving: \(A=-1, B=1, C=1\).
Final result: \(\dfrac{-1}{x+1} + \dfrac{x+1}{x^2+1}\).
Binomial Series
Binomial Series
For rational \(n\) and \(|x|<1\), the binomial expansion of \((1+x)^n\) is
\((1+x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \cdots\),
valid for \(|x|<1\). (You are not required to work with the general term; use the first few terms as needed.)
If the expression is of the form \((a + bx)^n\) with \(a\neq 0\), rewrite as
\((a+bx)^n = a^n\Big(1 + \dfrac{b}{a}x\Big)^n\),
and apply the binomial series to \(\Big(1 + \dfrac{b}{a}x\Big)^n\). The condition for validity becomes \(\Big|\dfrac{b}{a}x\Big|<1\), i.e. \(|x|<\dfrac{|a|}{|b|}\).
Example
Expand \((1+x)^{\tfrac{1}{2}}\) up to and including the \(x^3\)-term, and state the range of \(x\) for which the expansion is valid.
▶️ Answer / Explanation
Use the binomial series with \(n=\tfrac{1}{2}\):
\((1+x)^{\tfrac{1}{2}} = 1 + \dfrac{1}{2}x + \dfrac{\tfrac{1}{2}(\tfrac{1}{2}-1)}{2!}x^2 + \dfrac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!}x^3 + \cdots\)
Compute coefficients:
\(\dfrac{1}{2}(\tfrac{1}{2}-1)=\dfrac{1}{2}\cdot(-\tfrac{1}{2})=-\dfrac{1}{4}\), so the \(x^2\)-coefficient is \(-\dfrac{1}{8}\).
For the \(x^3\)-coefficient: \(\dfrac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!} = \dfrac{\tfrac{1}{2}\cdot(-\tfrac{1}{2})\cdot(-\tfrac{3}{2})}{6} = \dfrac{3}{48} = \dfrac{1}{16}.\)
Thus, up to \(x^3\):
\(\displaystyle (1+x)^{\tfrac{1}{2}} \approx 1 + \dfrac{1}{2}x – \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3.\)
Validity: the expansion holds for \(|x|<1\).
Final boxed result:
\(\boxed{(1+x)^{\tfrac{1}{2}} \approx 1 + \dfrac{1}{2}x – \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3,\quad |x|<1.}\)
Example
Expand \(\displaystyle \bigl(2 – \tfrac{x}{2}\bigr)^{-1}\) in ascending powers of \(x\) (up to \(x^3\)), and determine the values of \(x\) for which the expansion is valid.
▶️ Answer / Explanation
First rewrite to the standard form. Factor out \(2\):
\(\displaystyle \bigl(2 – \tfrac{x}{2}\bigr)^{-1} = 2^{-1}\Big(1 – \dfrac{x}{4}\Big)^{-1} = \dfrac{1}{2}\Big(1 – \dfrac{x}{4}\Big)^{-1}.\)
Now use the binomial series with \(n=-1\) (this is the geometric series):
\p style=”padding-left: 40px;”>\(\displaystyle (1+u)^{-1} = 1 – u + u^2 – u^3 + \cdots\) for \(|u|<1\). Here \(u = -\dfrac{x}{4}\).
Substitute \(u = -\dfrac{x}{4}\):
\(\displaystyle \Big(1 – \dfrac{x}{4}\Big)^{-1} = 1 + \dfrac{x}{4} + \dfrac{x^2}{16} + \dfrac{x^3}{64} + \cdots\)
Multiply by \(\dfrac{1}{2}\):
\(\displaystyle \bigl(2 – \tfrac{x}{2}\bigr)^{-1} = \dfrac{1}{2}\Big(1 + \dfrac{x}{4} + \dfrac{x^2}{16} + \dfrac{x^3}{64} + \cdots\Big)\)
So up to \(x^3\):
\(\displaystyle \bigl(2 – \tfrac{x}{2}\bigr)^{-1} \approx \dfrac{1}{2} + \dfrac{x}{8} + \dfrac{x^2}{32} + \dfrac{x^3}{128}.\)
Validity condition: \(\displaystyle \Big|-\dfrac{x}{4}\Big|<1 \Longrightarrow |x|<4.\)
Final boxed result:
\(\boxed{\displaystyle \bigl(2 – \tfrac{x}{2}\bigr)^{-1} \approx \dfrac{1}{2} + \dfrac{x}{8} + \dfrac{x^2}{32} + \dfrac{x^3}{128},\quad |x|<4.}\)
General remark on validity (useful checklist)
- If you rewrite \((a+bx)^n = a^n\bigl(1 + \tfrac{b}{a}x\bigr)^n\), the necessary condition for the binomial expansion is \(\bigl|\tfrac{b}{a}x\bigr|<1\), i.e. \(|x|<\tfrac{|a|}{|b|}\).
- For geometric-type expansions \((1+u)^{-1}\) you use \(|u|<1\); for other rational \(n\) the same \(|u|<1\) requirement applies for convergence of the power series about \(u=0\).