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SingaporeMathsSyllabus dot point

How do we expand, factorise and simplify algebraic expressions reliably?

Expand products, factorise expressions including quadratics and the difference of two squares, and simplify algebraic fractions

A focused answer to the O-Level E-Maths outcome on algebraic manipulation. Expanding brackets, factorising by common factor, grouping, the difference of two squares and quadratics, and simplifying algebraic fractions.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to expand products of brackets, factorise a range of expressions (common factors, grouping, the difference of two squares, and quadratic trinomials), and simplify algebraic fractions. Factorising is the reverse of expanding, and it is the key that unlocks solving quadratics and simplifying fractions later.

The answer

Expanding brackets

To expand a single bracket, multiply each term inside by the term outside. To expand two brackets, multiply every term in the first by every term in the second:

(x+3)(xβˆ’5)=x2βˆ’5x+3xβˆ’15=x2βˆ’2xβˆ’15(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15

Useful identities are (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 and (aβˆ’b)(a+b)=a2βˆ’b2(a - b)(a + b) = a^2 - b^2.

Factorising by common factor

Take out the highest common factor of all terms: 6x2+9x=3x(2x+3)6x^2 + 9x = 3x(2x + 3). Always check first whether a common factor exists, because it simplifies everything that follows.

Factorising by grouping

When four terms share factors in pairs, group and factorise each pair, then take out the common bracket:

ax+ay+bx+by=a(x+y)+b(x+y)=(a+b)(x+y)ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)

The difference of two squares

An expression of the form a2βˆ’b2a^2 - b^2 factorises as (aβˆ’b)(a+b)(a - b)(a + b). Spotting this pattern, including with coefficients such as 9x2βˆ’16=(3xβˆ’4)(3x+4)9x^2 - 16 = (3x - 4)(3x + 4), is a frequent exam shortcut.

Factorising quadratics

For x2+bx+cx^2 + bx + c, find two numbers that multiply to cc and add to bb. For x2+5x+6x^2 + 5x + 6 the numbers are 22 and 33, giving (x+2)(x+3)(x + 2)(x + 3). When the coefficient of x2x^2 is not 11, split the middle term using factors that multiply to give the product of the outer coefficients.

Simplifying algebraic fractions

Factorise the numerator and denominator fully, then cancel any common factors. You can only cancel a factor that multiplies the whole top and the whole bottom, never an individual term.

Examples in context

Example 1. Solving equations. Factorising x2βˆ’2xβˆ’15=0x^2 - 2x - 15 = 0 into (xβˆ’5)(x+3)=0(x - 5)(x + 3) = 0 immediately gives the solutions x=5x = 5 and x=βˆ’3x = -3. Factorisation is the bridge from an expression to the roots of an equation.

Example 2. Simplifying a formula. A rectangle of length x+2x + 2 and width xx has area x(x+2)=x2+2xx(x + 2) = x^2 + 2x. Expanding and factorising let you move between a factorised geometric form and an expanded algebraic one as a problem requires.

Try this

Q1. Expand and simplify (xβˆ’4)(x+6)(x - 4)(x + 6). [2 marks]

  • Cue. x2+6xβˆ’4xβˆ’24=x2+2xβˆ’24x^2 + 6x - 4x - 24 = x^2 + 2x - 24.

Q2. Factorise 25βˆ’16y225 - 16y^2. [1 mark]

  • Cue. Difference of two squares: (5βˆ’4y)(5+4y)(5 - 4y)(5 + 4y).

Q3. Factorise x2βˆ’3xβˆ’10x^2 - 3x - 10. [2 marks]

  • Cue. Two numbers multiplying to βˆ’10-10 and adding to βˆ’3-3 are βˆ’5-5 and 22, giving (xβˆ’5)(x+2)(x - 5)(x + 2).

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksFactorise completely 3x2βˆ’123x^2 - 12.
Show worked answer β†’

First take out the common factor 33: 3x2βˆ’12=3(x2βˆ’4)3x^2 - 12 = 3(x^2 - 4).

The bracket is a difference of two squares: x2βˆ’4=(xβˆ’2)(x+2)x^2 - 4 = (x - 2)(x + 2).

So 3x2βˆ’12=3(xβˆ’2)(x+2)3x^2 - 12 = 3(x - 2)(x + 2).

Markers reward extracting the common factor first, recognising the difference of two squares, and the fully factorised form.

Original4 marksSimplify x2βˆ’9x2+7x+12\dfrac{x^2 - 9}{x^2 + 7x + 12}.
Show worked answer β†’

Factorise the numerator as a difference of two squares: x2βˆ’9=(xβˆ’3)(x+3)x^2 - 9 = (x - 3)(x + 3).

Factorise the denominator: x2+7x+12=(x+3)(x+4)x^2 + 7x + 12 = (x + 3)(x + 4), since 3+4=73 + 4 = 7 and 3Γ—4=123 \times 4 = 12.

Cancel the common factor (x+3)(x + 3): (xβˆ’3)(x+3)(x+3)(x+4)=xβˆ’3x+4\dfrac{(x - 3)(x + 3)}{(x + 3)(x + 4)} = \dfrac{x - 3}{x + 4}.

Markers reward factorising both numerator and denominator and cancelling the common bracket to reach the simplified fraction.

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