IB MYP 3 Mathematics 3.2 Expanding, Factorising and Algebraic Fractions Study Notes - New Syllabus
IB MYP 3 Mathematics 3.2 Expanding, Factorising and Algebraic Fractions Study Notes
IB MYP 3 Mathematics 3.2 Expanding, Factorising and Algebraic Fractions Study Notes at IITian Academy focus on specific topics and types of questions asked in the actual exam. Study Notes focus on the IB MYP 3 Mathematics syllabus with guiding questions of
Distributive law: \(a(b+c)=ab+ac\).
Expansion: Remove brackets by multiplying each term correctly.
Common factor: The greatest factor shared by all terms can be taken outside a bracket.
Quadratic factorisation: Find factors whose product and sum match the required coefficients.
Difference of squares: \(a^2-b^2=(a-b)(a+b)\).
Algebraic fraction: A fraction containing variables.
Simplifying algebraic fractions: Factorise first, then cancel common factors.
Adding/subtracting: Use a common denominator.
Multiplying: Multiply numerators and denominators, then simplify.
Dividing: Multiply by the reciprocal.
Restriction: A denominator can never equal zero.
3.2 – Expanding, Factorising and Algebraic Fractions
Expanding and factorising are opposite algebraic processes. Expanding removes brackets, while factorising puts an expression into a product of factors. Algebraic fractions extend these skills to expressions involving variables in the numerator and denominator.
Expansion and Factorisation
Expansion and factorisation are inverse processes.
\(3(x+4)=3x+12\)
Therefore, the reverse process is:
\(3x+12=3(x+4)\)
Expansion → brackets are removed.
Factorisation → brackets are introduced.
Expanding a Single Bracket
Use the distributive law to multiply the term outside the bracket by every term inside.
\(a(b+c)=ab+ac\)
Example:
\(5(2x-3)\)
\(=10x-15\)
With a negative term:
\(-3(2x-5)\)
\(=-6x+15\)
The number outside the bracket must be multiplied by every term inside the bracket.
Expanding Two Brackets
When multiplying two brackets, every term in the first bracket is multiplied by every term in the second bracket.
\((a+b)(c+d)=ac+ad+bc+bd\)
Example:
\((x+3)(x+5)\)
\(=x^2+5x+3x+15\)
\(=x^2+8x+15\)
With negative signs:
\((x-4)(x+2)\)
\(=x^2+2x-4x-8\)
\(=x^2-2x-8\)
² Special Expansion Identities
Some common expressions can be expanded using standard identities.
\((a+b)^2=a^2+2ab+b^2\)
\((a-b)^2=a^2-2ab+b^2\)
For example:
\((x+4)^2\)
\(=x^2+8x+16\)
Difference of Two Squares
\(a^2-b^2=(a-b)(a+b)\)
Example:
\(x^2-25=(x-5)(x+5)\)
Factorising by Taking Out a Common Factor
To factorise an expression, first look for the greatest common factor shared by all terms.
Example:
\(12x+18\)
The greatest common factor is \(6\).
\(12x+18=6(2x+3)\)
With variables:
\(15x^2+10x\)
\(=5x(3x+2)\)
Expand your factorised answer. If you obtain the original expression, your factorisation is correct.
Factorising Quadratic Expressions
For a quadratic expression of the form:
\(x^2+bx+c\)
find two numbers whose:
- product is \(c\)
- sum is \(b\)
Example:
\(x^2+7x+12\)
The two numbers are \(3\) and \(4\), because:
\(3\times4=12\)
\(3+4=7\)
Therefore:
\(x^2+7x+12=(x+3)(x+4)\)
Algebraic Fractions
An algebraic fraction is a fraction containing one or more variables.
Examples include:
- \(\dfrac{x}{5}\)
- \(\dfrac{3x}{4y}\)
- \(\dfrac{x+2}{x-3}\)
The denominator of an algebraic fraction cannot equal zero.
For example:
\(\dfrac{5}{x-2}\)
Here \(x\neq2\), because \(x-2=0\) when \(x=2\).
Simplifying Algebraic Fractions
To simplify an algebraic fraction, factorise the numerator and denominator and cancel common factors.
Example:
\(\dfrac{x^2+5x}{x}\)
Factorise the numerator:
\(=\dfrac{x(x+5)}{x}\)
Cancel the common factor \(x\):
\(=x+5\), where \(x\neq0\)
You can only cancel factors, not terms being added or subtracted.
For example, \(\dfrac{x+5}{x}\) cannot be simplified by cancelling the \(x\).
Adding and Subtracting Algebraic Fractions
Algebraic fractions must have a common denominator before they can be added or subtracted.
If the denominator is the same:
\(\dfrac{a}{c}+\dfrac{b}{c}=\dfrac{a+b}{c}\)
Example:
\(\dfrac{3x}{5}+\dfrac{2x}{5}=\dfrac{5x}{5}=x\)
With different denominators, find a common denominator.
\(\dfrac{x}{3}+\dfrac{x}{4}\)
\(=\dfrac{4x}{12}+\dfrac{3x}{12}\)
\(=\dfrac{7x}{12}\)
Multiplying Algebraic Fractions
Multiply the numerators together and denominators together, then simplify.
\(\dfrac{a}{b}\times\dfrac{c}{d}=\dfrac{ac}{bd}\)
Example:
\(\dfrac{3x}{4}\times\dfrac{8}{9x}\)
Cancel common factors before multiplying:
\(=\dfrac{3x\times8}{4\times9x}\)
\(=\dfrac{2}{3}\), where \(x\neq0\)
Dividing Algebraic Fractions
To divide algebraic fractions, multiply by the reciprocal of the second fraction.
Example:
\(\dfrac{x}{3}\div\dfrac{2x}{9}\)
\(=\dfrac{x}{3}\times\dfrac{9}{2x}\)
\(=\dfrac{3}{2}\), where \(x\neq0\)
1. Factorise where possible.
2. Cancel common factors only.
3. For addition/subtraction, use a common denominator.
4. For multiplication, multiply numerators and denominators.
5. For division, multiply by the reciprocal.
6. State restrictions when a variable could make a denominator zero.
Example 1:
Let
\(A=(x+4)(x-3)\)
a) Expand \(A\).
b) Factorise \(x^2+7x+12\).
c) Factorise \(2x^2+8x\).
d) Factorise \(x^2-36\).
▶️ Answer/Explanation
Answer
a)
\(A=(x+4)(x-3)\)
\(=x^2-3x+4x-12\)
\(=x^2+x-12\)
Answer: \(x^2+x-12\)
b)
Two numbers with product \(12\) and sum \(7\) are \(3\) and \(4\).
\(x^2+7x+12=(x+3)(x+4)\)
Answer: \((x+3)(x+4)\)
c)
The common factor is \(2x\).
\(2x^2+8x=2x(x+4)\)
Answer: \(2x(x+4)\)
d)
\(x^2-36=x^2-6^2\)
\(=(x-6)(x+6)\)
Answer: \((x-6)(x+6)\)
Example 2:
Simplify each expression. State any values of the variable that are not allowed.
a) \(\dfrac{x^2+6x}{x}\)
b) \(\dfrac{x}{3}+\dfrac{x}{6}\)
c) \(\dfrac{4x}{5}\times\dfrac{10}{8x}\)
d) \(\dfrac{x}{4}\div\dfrac{3x}{8}\)
▶️ Answer/Explanation
Answer
a)
Factorise the numerator:
\(\dfrac{x(x+6)}{x}\)
\(=x+6\), where \(x\neq0\)
Answer: \(x+6,\;x\neq0\)
b)
The common denominator is \(6\).
\(\dfrac{x}{3}+\dfrac{x}{6}=\dfrac{2x}{6}+\dfrac{x}{6}\)
\(=\dfrac{3x}{6}\)
\(=\dfrac{x}{2}\)
Answer: \(\dfrac{x}{2}\)
c)
\(\dfrac{4x}{5}\times\dfrac{10}{8x}\)
\(=\dfrac{40x}{40x}=1\)
Answer: \(1,\;x\neq0\)
d)
\(\dfrac{x}{4}\div\dfrac{3x}{8}\)
\(=\dfrac{x}{4}\times\dfrac{8}{3x}\)
\(=\dfrac{2}{3}\)
Answer: \(\dfrac{2}{3},\;x\neq0\)