Edexcel Mathematics (4XMAH) -Unit 1 - 2.2 Algebraic Manipulation- Study Notes- New Syllabus
Edexcel Mathematics (4XMAH) -Unit 1 – 2.2 Algebraic Manipulation- Study Notes- New syllabus
Edexcel Mathematics (4XMAH) -Unit 1 – 2.2 Algebraic Manipulation- Study Notes -Edexcel iGCSE Mathematics – per latest Syllabus.
Key Concepts:
A expand the product of two or more linear expressions
Expand and simplify: (x + 2)(x + 3)(x − 1)
B understand the concept of a quadratic expression and be able to factorise such expressions
Factorise: 6x² − 5x − 6
C manipulate algebraic fractions where the numerator and/or the denominator can be numeric, linear or quadratic
Express as a single fraction:
3x + 1/(x + 2) − (x − 2)/(x − 1)
Simplify: (2x² + 3x)/(4x² − 9)
D complete the square for a given quadratic expression
Write 2x² + 6x − 1 in the form a(x + b)² + c
Edexcel iGCSE Mathematics -Concise Summary Notes- All Topics
Expanding Linear Expressions
Expanding means multiplying brackets to remove them and simplify the expression.
We use the distributive law:
\( a(b+c)=ab+ac \)
Two Brackets
Multiply every term in the first bracket by every term in the second bracket.
\( (a+b)(c+d)=ac+ad+bc+bd \)
After expanding, always collect like terms.
Key Idea
Each term multiplies every term in the other bracket.
Example 1:
Expand \( (x+3)(x+5) \).
▶️ Answer/Explanation
\( x(x+5)+3(x+5) \)
\( =x^2+5x+3x+15 \)
\( =x^2+8x+15 \)
Conclusion: \( x^2+8x+15 \).
Example 2:
Expand and simplify \( (2x+1)(x+4) \).
▶️ Answer/Explanation
\( 2x(x+4)+1(x+4) \)
\( =2x^2+8x+x+4 \)
\( =2x^2+9x+4 \)
Conclusion: \( 2x^2+9x+4 \).
Example 3:
Expand and simplify \( (x+y)(x+1) \).
▶️ Answer/Explanation
Multiply each term:
\( x(x+1)+y(x+1) \)
\( =x^2+x+xy+y \)
Conclusion: \( x^2+xy+x+y \).
Quadratic Expressions and Factorisation
A quadratic expression is an algebraic expression where the highest power of \( x \) is 2.
\( ax^2+bx+c \)
Examples:
\( x^2+5x+6 \)
\( 3x^2-2x+1 \)
Factorising a Quadratic
Factorising means writing the quadratic as a product of two brackets.
\( ax^2+bx+c=(px+q)(rx+s) \)
Method (AC Method)
1. Multiply \( a \times c \).
2. Find two numbers that multiply to this value and add to \( b \).
3. Split the middle term.
4. Factorise by grouping.
Example 1:
Factorise \( x^2+7x+10 \).
▶️ Answer/Explanation
Numbers that multiply to 10 and add to 7 are 5 and 2.
\( (x+5)(x+2) \)
Conclusion: \( (x+5)(x+2) \).
Example 2:
Factorise \( 2x^2+9x+4 \).
▶️ Answer/Explanation
\( a\times c=2\times4=8 \)
Numbers: 8 and 1 (sum 9)
\( 2x^2+8x+x+4 \)
\( 2x(x+4)+1(x+4) \)
\( (2x+1)(x+4) \)
Conclusion: \( (2x+1)(x+4) \).
Example 3:
Factorise \( 6x^2-5x-6 \).
▶️ Answer/Explanation
\( a\times c=6\times(-6)=-36 \)
Numbers: -9 and 4 (sum -5)
\( 6x^2-9x+4x-6 \)
\( 3x(2x-3)+2(2x-3) \)
\( (3x+2)(2x-3) \)
Conclusion: \( (3x+2)(2x-3) \).
Algebraic Fractions
An algebraic fraction is a fraction where the numerator or denominator contains algebraic expressions.
\( \dfrac{x+2}{x-1} \)
Adding or Subtracting Algebraic Fractions
Fractions can only be added when they have a common denominator.
Find the lowest common denominator (LCD), then combine the numerators.
\( \dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd} \)
Simplifying Algebraic Fractions
Factorise the numerator and denominator, then cancel common factors.
\( \dfrac{ab}{ac}=\dfrac{b}{c} \)
Important
You may only cancel factors, not terms.
Example 1:
Express as a single fraction
\( \dfrac{3x+1}{x+2}+\dfrac{x-2}{x-1} \)
▶️ Answer/Explanation
Common denominator:
\( (x+2)(x-1) \)
Combine:
\( \dfrac{(3x+1)(x-1)+(x-2)(x+2)}{(x+2)(x-1)} \)
\( =\dfrac{(3x^2-3x+x-1)+(x^2-4)}{(x+2)(x-1)} \)
\( =\dfrac{3x^2-2x-1+x^2-4}{(x+2)(x-1)} \)
\( =\dfrac{4x^2-2x-5}{(x+2)(x-1)} \)
Conclusion: \( \dfrac{4x^2-2x-5}{(x+2)(x-1)} \).
Example 2:
Simplify \( \dfrac{2x^2+3x}{4x^2-9} \).
▶️ Answer/Explanation
Factorise:
\( 2x^2+3x=x(2x+3) \)
\( 4x^2-9=(2x-3)(2x+3) \)
Cancel common factor:
\( \dfrac{x(2x+3)}{(2x-3)(2x+3)}=\dfrac{x}{2x-3} \)
Conclusion: \( \dfrac{x}{2x-3} \).
Example 3:
Express as a single fraction
\( \dfrac{1}{x}+\dfrac{2}{x+1} \)
▶️ Answer/Explanation
Common denominator:
\( x(x+1) \)
\( \dfrac{x+1}{x(x+1)}+\dfrac{2x}{x(x+1)} \)
\( =\dfrac{x+1+2x}{x(x+1)}=\dfrac{3x+1}{x(x+1)} \)
Conclusion: \( \dfrac{3x+1}{x(x+1)} \).
Completing the Square
Completing the square is a method used to rewrite a quadratic expression in the form
\( a(x+b)^2+c \)
This form is useful for solving equations and identifying turning points of graphs.
Steps
1. Factorise the coefficient of \( x^2 \) from the first two terms.
2. Take half of the coefficient of \( x \).
3. Square it and add inside the bracket.
4. Adjust the constant outside the bracket.
Key Idea
\( x^2+2bx+b^2=(x+b)^2 \)
Example 1:
Write \( x^2+6x+5 \) in the form \( (x+b)^2+c \).
▶️ Answer/Explanation
Half of 6 is 3.
\( (x+3)^2=x^2+6x+9 \)
\( x^2+6x+5=(x+3)^2-4 \)
Conclusion: \( (x+3)^2-4 \).
Example 2:
Write \( x^2-4x+1 \) in the form \( (x+b)^2+c \).
▶️ Answer/Explanation
Half of −4 is −2.
\( (x-2)^2=x^2-4x+4 \)
\( x^2-4x+1=(x-2)^2-3 \)
Conclusion: \( (x-2)^2-3 \).
Example 3:
Write \( 2x^2+6x-1 \) in the form \( a(x+b)^2+c \).
▶️ Answer/Explanation
Factorise 2 from first two terms:
\( 2(x^2+3x)-1 \)
Half of 3 is \( \dfrac{3}{2} \), square it:
\( \left(\dfrac{3}{2}\right)^2=\dfrac{9}{4} \)
\( 2\left(x^2+3x+\dfrac{9}{4}\right)-2\cdot\dfrac{9}{4}-1 \)
\( 2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}-1 \)
\( 2\left(x+\dfrac{3}{2}\right)^2-\dfrac{11}{2} \)
Conclusion: \( 2\left(x+\dfrac{3}{2}\right)^2-\dfrac{11}{2} \).