This article covers the essential skills of algebraic fractions and factorization – a core topic in DSE Paper 1 Section A(1) and A(2) worth 4–6 marks. You will learn how to simplify algebraic fractions, perform the four operations (addition, subtraction, multiplication, division), and apply factorization techniques including common factor, grouping, cross-method, and identities. The article also introduces the Factor Theorem and Remainder Theorem – key tools for polynomial manipulation. Includes step-by-step worked examples, DSE exam techniques, practice questions, and a full set of homework with solutions. These are essential skills that form the foundation for more advanced algebra topics.
By the end of this article, you should be able to:
Algebraic fractions and factorization appear every year in DSE Paper 1 Section A(1) and A(2). Questions may ask you to:
These are essential skills that also appear in more advanced topics like quadratic equations and functions. Master them early!
Questions on algebraic fractions and factorization often appear in Section A(1) and A(2) as short-answer questions worth 3–4 marks each. They test your algebraic manipulation skills.
An algebraic fraction is a fraction where the numerator and/or denominator contain algebraic expressions. To simplify, factorize both the numerator and denominator, then cancel common factors.
Simplify: \( \frac{x^2 - 4}{x^2 - 2x} \)
\( \frac{x^2 - 4}{x^2 - 2x} = \frac{(x-2)(x+2)}{x(x-2)} = \frac{x+2}{x} \), where \( x \neq 0, 2 \)
You cannot cancel terms that are added or subtracted – only factors that are multiplied. For example, you cannot cancel \( x \) in \( \frac{x+2}{x+3} \).
$$ \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd} $$
Factorize and cancel first, then multiply.
\( \frac{x^2-1}{x} \times \frac{2x}{x+1} = \frac{(x-1)(x+1)}{x} \times \frac{2x}{x+1} = 2(x-1) \)
$$ \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc} $$
Flip the second fraction (reciprocal), then multiply.
\( \frac{x^2-9}{x} \div \frac{x-3}{x+2} = \frac{(x-3)(x+3)}{x} \times \frac{x+2}{x-3} = \frac{(x+3)(x+2)}{x} \)
$$ \frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd} $$
Find the lowest common denominator (LCD), then combine numerators.
\( \frac{2}{x} + \frac{3}{x+1} = \frac{2(x+1) + 3x}{x(x+1)} = \frac{2x+2+3x}{x(x+1)} = \frac{5x+2}{x(x+1)} \)
Identify the highest common factor (HCF) of all terms and factor it out.
\( 6x^2 + 9x = 3x(2x + 3) \)
Group terms with common factors, then factor each group.
\( 2x^2 + 6x + 3x + 9 = 2x(x+3) + 3(x+3) = (2x+3)(x+3) \)
For \( ax^2 + bx + c \), find two numbers that multiply to \( ac \) and add to \( b \).
\( 2x^2 + 7x + 3 = (2x+1)(x+3) \)
Check: \( (2x+1)(x+3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3 \)
\( 9x^2 - 16 = (3x-4)(3x+4) \)
\( x^2 + 6x + 9 = (x+3)^2 \)
If \( P(x) \) is a polynomial and \( P(a) = 0 \), then \( (x-a) \) is a factor of \( P(x) \).
Factor Theorem: If \( P(a) = 0 \), then \( (x-a) \) is a factor.
Question: Show that \( (x-2) \) is a factor of \( P(x) = x^3 - 3x^2 + 4 \).
\( P(2) = 2^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0 \)
Since \( P(2) = 0 \), \( (x-2) \) is a factor of \( P(x) \).
When a polynomial \( P(x) \) is divided by \( (x-a) \), the remainder is \( P(a) \).
Remainder Theorem: The remainder when dividing \( P(x) \) by \( (x-a) \) is \( P(a) \).
Question: Find the remainder when \( P(x) = 2x^3 - 3x + 5 \) is divided by \( (x-1) \).
\( P(1) = 2(1)^3 - 3(1) + 5 = 2 - 3 + 5 = 4 \)
Remainder = 4
Question: Simplify \( \frac{x^2 - 5x + 6}{x^2 - 9} \).
Factorize numerator: \( x^2 - 5x + 6 = (x-2)(x-3) \)
Factorize denominator: \( x^2 - 9 = (x-3)(x+3) \)
\( \frac{(x-2)(x-3)}{(x-3)(x+3)} = \frac{x-2}{x+3} \), where \( x \neq \pm 3 \)
Question: Simplify \( \frac{3}{x-1} + \frac{2}{x+2} \).
LCD = \( (x-1)(x+2) \)
\( \frac{3(x+2) + 2(x-1)}{(x-1)(x+2)} = \frac{3x+6+2x-2}{(x-1)(x+2)} = \frac{5x+4}{(x-1)(x+2)} \)
Question: Factorize \( P(x) = x^3 - 4x^2 + x + 6 \) completely.
Test \( x = 2 \): \( P(2) = 8 - 16 + 2 + 6 = 0 \). So \( (x-2) \) is a factor.
Divide: \( P(x) = (x-2)(x^2 - 2x - 3) \)
Factorize quadratic: \( x^2 - 2x - 3 = (x-3)(x+1) \)
Answer: \( P(x) = (x-2)(x-3)(x+1) \)
Question 1 MC
Simplify \( \frac{x^2 - 4x + 4}{x^2 - 4} \).
A. \( \frac{x-2}{x+2} \) B. \( \frac{x+2}{x-2} \) C. \( \frac{x-2}{x} \) D. \( \frac{x}{x+2} \)
Question 2 MC
Factorize \( 2x^2 + 5x - 3 \).
A. \( (2x-1)(x+3) \) B. \( (2x+1)(x-3) \) C. \( (2x-3)(x+1) \) D. \( (2x+3)(x-1) \)
Question 3 Short Answer
Simplify \( \frac{2}{x} + \frac{3}{x-1} \).
Question 4 Short Answer
Find the remainder when \( P(x) = x^3 - 2x^2 + 3x - 4 \) is divided by \( (x-2) \).
Question 5 MC
Which of the following is a factor of \( P(x) = x^3 - 3x^2 - 4x + 12 \)?
A. \( (x-2) \) B. \( (x+2) \) C. \( (x-3) \) D. \( (x+3) \)
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An Exercise on Algebraic Fractions and Factorization | Complete Guide