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Sec 4 E Maths: Algebraic Expansion, Factorisation and Fractions, practice questions & worked solutions

Algebra questions selected from 2023 to 2025 Singapore school E Maths prelim papers, each with the school’s full worked solution.

12 questionsStep-by-step solutionsFree to access

About this topic & key methods

Expansion, factorisation and algebraic fractions are taught across Sec 1 to Sec 3 and run through almost every E Maths paper. Most questions here are short Paper 1 items, with two Paper 2 questions that combine several skills.

Attempt each question on paper before opening its worked solution. The solutions are the schools’ own marking schemes, so some steps are brief.

Key methods for factorisation and algebraic fractions

Expand term by term and collect like terms, watching the signs inside a subtracted bracket
Factorise in order: common factor first, then grouping in pairs, then $a^2-b^2=(a+b)(a-b)$ or a trinomial
Use identities such as $(a-b)^2=a^2-2ab+b^2$ to find values without solving
Factorise numerator and denominator fully before cancelling a common factor
Rewrite $1-2a$ as $-(2a-1)$ so that fractions share a denominator
To change the subject, clear fractions, collect the new subject on one side, factorise it out, then divide
To prove a multiple of $k$ , expand and simplify, then write the result as $k\times$ (an integer)

For lessons that build these methods week by week, see our E Maths tuition for Sec 3 and Sec 4.

Algebraic Expansion, Factorisation and Fractions questions with worked solutions

Unless the question specifies otherwise, give non-exact answers to 3 significant figures and angles in degrees to 1 decimal place.

Q1Yio Chu Kang Secondary School2025Prelim P1 Q22●○○ Foundational2 marks

Expanding a quadratic by a linear factor, Yio Chu Kang Secondary School, 2025 Prelim P1 Q22

Expand $(2x^2-5x+3)(5x-4)$ . [2]

Show worked solution▾

$\begin{aligned}(2x^2-5x+3)(5x-4)&=10x^3-8x^2-25x^2+20x+15x-12\\&=10x^3-33x^2+35x-12\end{aligned}$

Answer: $10x^3-33x^2+35x-12$

Q2Canberra Secondary School2024Prelim P1 Q3●○○ Foundational2 marks

Subtracting fractions with opposite denominators, Canberra Secondary School, 2024 Prelim P1 Q3

Write as a single fraction in its simplest form $\dfrac3{2a-1}-\dfrac2{1-2a}$ . [2]

Show worked solution▾

$\begin{aligned}&\frac3{2a-1}-\frac2{1-2a}\\&=\frac3{2a-1}+\frac2{2a-1}\\&=\frac5{2a-1}\end{aligned}$

Answer: $\dfrac5{2a-1}$

Q3Presbyterian High School2025Prelim P1 Q17●●○ Standard2 marks

Using a perfect square identity, Presbyterian High School, 2025 Prelim P1 Q17

Given that $2x-y=8$ and $xy=4$ , without solving for $x$ and $y$ , find the value of $4x^2+y^2$ . [2]

Show worked solution▾

$\begin{aligned}(2x-y)^2&=8^2\\4x^2-4xy+y^2&=64\\4x^2+y^2&=64+4xy\\4x^2+y^2&=64+4(4)\\4x^2+y^2&=80\end{aligned}$

Answer: $80$

Q4Bendemeer Secondary School2024Prelim P1 Q9●●○ Standard2 marks

Proving an expression is a multiple of 3, Bendemeer Secondary School, 2024 Prelim P1 Q9

Explain why $(2n+3)^2-(4n+3)(n-6)$ is a multiple of 3 for all integer values of $n$ . [2]

Show worked solution▾

$\begin{aligned}&(2n+3)^2-(4n+3)(n-6)\\&=4n^2+12n+9-(4n^2+3n-24n-18)\\&=33n+27\\&=3(11n+9)\end{aligned}$

Hence, it is a multiple of 3 for all integer values of $n$ .

Answer: $33n+27=3(11n+9)$

Q5Chung Cheng High School (Yishun)2025Prelim P1 Q4●●○ Standard3 marks

Difference of two squares for mental arithmetic, Chung Cheng High School (Yishun), 2025 Prelim P1 Q4

Simplify $(4n)^2-(4n+a)(4n-a)$ . [1]
Hence, without the use of a calculator, evaluate $75^2-(78)(72)$ . [2]

Show worked solution▾

(a)

$=16n^2-[(4n)^2-a^2]$

$=a^2$

(b)

$=(75)^2-[(75+3)(75-3)]$

$=3^2$

$=9$

Answer: $\text{(a) }a^2\quad\text{(b) }9$

Q6Anglican High School2025Prelim P2 Q1●●○ Standard3 marks

Changing the subject inside a squared bracket, Anglican High School, 2025 Prelim P2 Q1

Rearrange the formula to make $c$ the subject. $D=(a-c)^2-a^2$ [3]

Show worked solution▾

$\begin{aligned}D&=(a-c)^2-a^2\\(a-c)^2&=D+a^2\\a-c&=\pm\sqrt{D+a^2}\\c&=a\pm\sqrt{D+a^2}\end{aligned}$

Answer: $c=a\pm\sqrt{D+a^2}$

Q7St Joseph’s Institution2025Prelim P1 Q10●●○ Standard3 marks

Subtracting algebraic fractions with a quadratic denominator, St Joseph’s Institution, 2025 Prelim P1 Q10

Write as a single fraction in its simplest form $\dfrac1{2-x}-\dfrac{3x+2}{x^2-7x+10}$ . [3]

Show worked solution▾

$\frac1{2-x}-\frac{3x+2}{x^2-7x+10}$

$=\frac{-1}{x-2}-\frac{3x+2}{(x-2)(x-5)}$

$=\frac{-x+5-3x-2}{(x-2)(x-5)}$

$=\frac{-4x+3}{(x-2)(x-5)}$

Answer: $\dfrac{-4x+3}{(x-2)(x-5)}$

Q8Ahmad Ibrahim Secondary School2023Prelim P1 Q11●●○ Standard3 marks

Simplifying a fraction by grouping and trinomial factors, Ahmad Ibrahim Secondary School, 2023 Prelim P1 Q11

Simplify $\dfrac{15am-4n-5m+12an}{8n^2+6mn-5m^2}$ . [3]

Show worked solution▾

$\begin{aligned}\dfrac{15am-4n-5m+12an}{8n^2+6mn-5m^2}&=\dfrac{5m(3a-1)+4n(3a-1)}{(4n+5m)(2n-m)}\\[6pt]&=\dfrac{(3a-1)(5m+4n)}{(4n+5m)(2n-m)}\\[6pt]&=\dfrac{3a-1}{2n-m}\end{aligned}$

Answer: $\dfrac{3a-1}{2n-m}$

Q9Anderson Secondary School2024Prelim P1 Q7●●○ Standard3 marks

Perfect square and difference of two squares, Anderson Secondary School, 2024 Prelim P1 Q7

Factorise $9x^2+24xy+16y^2$ completely. [1]
Hence factorise $144a^8-(9a^8+24a^4y+16y^2)$ completely. [2]

Show worked solution▾

(a)

$9x^2+24xy+16y^2$

$=(3x+4y)^2$

(b)

Let $x=a^4$ ,

$144a^8-(9a^8+24a^4y+16y^2)$

$=144x^2-(9x^2+24xy+16y^2)$

$=(12x)^2-(3x+4y)^2$

$=(12x+3x+4y)(12x-3x-4y)$

$=(15x+4y)(9x-4y)$

$=(15a^4+4y)(9a^4-4y)$

Answer: $\text{(a) }(3x+4y)^2\quad\text{(b) }(15a^4+4y)(9a^4-4y)$

Q10Fairfield Methodist School (Secondary)2023Prelim P1 Q17●●○ Standard5 marks

Factorising by grouping and repeated difference of squares, Fairfield Methodist School (Secondary), 2023 Prelim P1 Q17

Factorise completely

$3ax+16by-12ay-4bx$ , [2]
$3mn-243mn^5$ . [3]

Show worked solution▾

(a)

$\begin{aligned}&3ax+16by-12ay-4bx\\&=3ax-12ay-4bx+16by\\&=3a(x-4y)-4b(x-4y)\\&=(3a-4b)(x-4y)\text{ or }(4b-3a)(4y-x)\end{aligned}$

(b)

$\begin{aligned}&3mn-243mn^5\\&=3mn\left(1-81n^4\right)\\&=3mn\left(1+9n^2\right)\left(1-9n^2\right)\\&=3mn\left(1+9n^2\right)(1+3n)(1-3n)\end{aligned}$

Answer: (a) $(3a-4b)(x-4y)$ or $(4b-3a)(4y-x)$ (b) $3mn\left(1+9n^2\right)(1+3n)(1-3n)$

Q11Anglo-Chinese School (Barker Road)2023Prelim P1 Q19●●● Challenging7 marks

Spotting a factorisation pattern to sum a long expression, Anglo-Chinese School (Barker Road), 2023 Prelim P1 Q19

1. By factorisation, show that $1+x+x(1+x)+x(1+x)^2=(1+x)^3$ . [2]
2. Given $1+x+x(1+x)+x(1+x)^2+x(1+x)^3+x(1+x)^4=(1+x)^n$ ,
  State the value of $n$ . [1]
Hence, evaluate
$2+2x+2x(1+x)+2x(1+x)^2+2x(1+x)^3+2x(1+x)^4+\ldots+2x(1+x)^{99}$
if $x=3$ .
Express your answer in index notation as a power of 2. [4]

Show worked solution▾

(a)(i)

$\begin{aligned}&(1+x)\left(1+x+x+x^2\right)\\&=(1+x)(1+x)^2\\&=(1+x)^3\end{aligned}$

(a)(ii)

$n=5$

(b)

$\begin{aligned}&2\left[(1+x)^{100}\right]\\&=2(4)^{100}\\&=2\left(2^{200}\right)\\&=2^{201}\end{aligned}$

Answer: (a)(i) shown (ii) $n=5$ (b) $2^{201}$

Q12Nan Chiau High School2024Prelim P2 Q1●●● Challenging9 marks

Algebraic fractions and a compound inequality, Nan Chiau High School, 2024 Prelim P2 Q1

Simplify $\dfrac{9a^2-4b^2}{3ap+6aq-4bq-2bp}$ . [3]
Write as a single fraction in its simplest form $\dfrac{x}{6-7x-5x^2}+\dfrac2{5x-3}$ . [3]
Solve the inequality $-\dfrac12<\dfrac{2y}5-\dfrac{1+y}3\le\dfrac56$ . [3]

Show worked solution▾

(a)

$\begin{aligned}\frac{9a^2-4b^2}{3ap+6aq-4bq-2bp}&=\frac{(3a+2b)(3a-2b)}{(3ap+6aq)-(2bp+4bq)}\\&=\frac{(3a+2b)(3a-2b)}{3a(p+2q)-2b(p+2q)}\\&=\frac{(3a+2b)(3a-2b)}{(p+2q)(3a-2b)}\\&=\frac{3a+2b}{p+2q}\end{aligned}$

(b)

$\begin{aligned}\frac x{6-7x-5x^2}+\frac2{5x-3}&=\frac x{-(5x-3)(x+2)}+\frac2{5x-3}\\&=\frac{-x+2(2+x)}{(5x-3)(x+2)}\\&=\frac{-x+4+2x}{(5x-3)(x+2)}\\&=\frac{x+4}{(5x-3)(x+2)}\end{aligned}$

(c)

$\begin{aligned}-\frac12&<\frac{2y}5-\frac{1+y}3\le\frac56\\-\frac12<\frac{3(2y)-5(1+y)}{15}\quad&\textit{and}\quad\frac{3(2y)-5(1+y)}{15}\le\frac56\\-\frac12<\frac{6y-5-5y}{15}\qquad&\qquad\frac{6y-5-5y}{15}\le\frac56\\-\frac12<\frac{y-5}{15}\qquad&\qquad\frac{y-5}{15}\le\frac56\\-15<2(y-5)\qquad&\qquad y-5\le\frac{25}2\\-15<2y-10\qquad&\qquad y\le17\tfrac12\\-5<2y\qquad&\\-2\tfrac12<y&\le17\tfrac12\end{aligned}$

Answer: (a) $\dfrac{3a+2b}{p+2q}$ (b) $\dfrac{x+4}{(5x-3)(x+2)}$ (c) $-2\frac12<y\le17\frac12$

Need help with factorisation and algebraic fractions? Our Sec 4 E Maths tuition classes work through prelim questions like these every week.

More practice related to factorisation and algebraic fractions

Sec 2 Maths: Expansion and Factorisation

Sec 2 Maths: Algebraic Fractions and Formulae

Sec 4 E Maths: Quadratic Equations and Completing the Square

Sec 4 E Maths: Equations and Inequalities

Sec 4 E Maths: Indices and Standard Form

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