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Cambridge IGCSE Mathematics — 0580 Extended

Topic 2.2: Algebra and graphs — Algebraic Manipulation

Collect like terms

Like terms have the same letters raised to the same powers. You can add or subtract their coefficients; unlike terms stay separate.

Method

  1. Underline (or rewrite in groups) every \(a^2\) term, every \(ab\) term, and every number.
  2. Add the coefficients in each group — a minus sign belongs to the term that follows it.
  3. Write one simplified term per group, in a consistent order (highest power first, then mixed terms, then constants).

Simplify \(2a^2 + 3ab - 1 + 5a^2 - 9ab + 4\).

Worked collection of like terms giving 7a squared minus 6ab plus 3
\(a^2\) terms \(2+5\), \(ab\) terms \(3-9\), numbers \(-1+4\): \(7a^2-6ab+3\).

Expand

Expanding means removing brackets by multiplication. Every term in one factor multiplies every term in the other.

Method

  1. Single bracket: multiply the outside term by each term inside.
  2. Two brackets: four products (first \(\times\) first, first \(\times\) second, second \(\times\) first, second \(\times\) second), then collect like terms.
  3. Three brackets: expand any two first, then multiply the result by the remaining bracket.

Expand \(3x(2x - 4y)\).

Worked expansion of 3x times 2x minus 4y equals 6x squared minus 12xy
\(3x \times 2x = 6x^2\) and \(3x \times (-4y) = -12xy\).

Expand \((3x + y)(x - 4y)\).

Worked expansion of 3x plus y times x minus 4y equals 3x squared minus 11xy minus 4y squared
Four products, then \(-12xy + xy = -11xy\): \(3x^2 - 11xy - 4y^2\).

Expand \((x - 2)(x + 3)(2x + 1)\).

Worked expansion of three brackets giving 2x cubed plus 3x squared minus 11x minus 6
Pair \((x-2)(x+3)=x^2+x-6\), then multiply by \(2x+1\).

Paper 2 (non-calculator)

For a product of three brackets, expand two of them first and simplify that quadratic before you multiply by the third. Carrying four or six uncollected terms into the last multiplication is how signs go missing.

Factorise fully

Factorising is expanding in reverse. Fully means every common factor — numbers and letters — has been taken out, and quadratics or cubics are written as products of linear factors where possible.

Method

  1. Take out the highest common factor of every term (number and letter).
  2. If four terms remain, try grouping into two pairs with the same bracket.
  3. Spot a difference of squares: \(A^2 - B^2 = (A - B)(A + B)\).
  4. For a quadratic \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\).

Common factor (number and letter):

\[9x^2 + 15xy = 3x(3x + 5y)\]

Factorise \(3x + 6y + ax + 2ay\) fully.

Worked factorisation by grouping giving 3 plus a times x plus 2y
Pairs \(3(x+2y)+a(x+2y)=(3+a)(x+2y)\).

Difference of squares:

\[9x^2 - 16y^2 = (3x - 4y)(3x + 4y)\]

Quadratic trinomial:

\[x^2 + 5x + 6 = (x + 2)(x + 3)\]

Factorise \(x^3 + 3x^2 + 2x\) fully.

Worked full factorisation of x cubed plus 3x squared plus 2x equals x times x plus 1 times x plus 2
First \(x\), then \(x^2+3x+2=(x+1)(x+2)\).

Completing the square

Completing the square rewrites a quadratic as a perfect square plus or minus a constant. Use it when the question asks for the form \((x+p)^2 + q\), or when you need the vertex of a parabola.

Method

  1. If the coefficient of \(x^2\) is not \(1\), factor that number from the \(x\)-terms only.
  2. Take half the coefficient of \(x\), then square it. Add and subtract that square inside the bracket.
  3. Rewrite as \(a(x+p)^2 + q\) and simplify \(q\).

Write \(x^2 + 6x + 5\) in completed-square form.

Worked completing the square x squared plus 6x plus 5 equals x plus 3 squared minus 4
Half of \(6\) is \(3\); \((x+3)^2 - 9 + 5 = (x+3)^2 - 4\).

Write \(2x^2 + 8x + 3\) in completed-square form.

Worked completing the square 2x squared plus 8x plus 3 equals 2 times x plus 2 squared minus 5
Factor \(2\) from the \(x\)-terms first: \(2(x+2)^2 - 5\).

Which method?

The question says expand, simplify, or remove brackets

Expand, then collect like terms.

The question says factorise (fully)

Common factor, grouping, difference of squares, or quadratic factors.

You need \((x+p)^2+q\), a vertex, or a turning point

Complete the square (factor \(a\) from the \(x\)-terms first if \(a \neq 1\)).

Try this

Factorise \(x^2 - 7x + 12\).

Show answer
Answer

  1. Two numbers that multiply to \(12\) and add to \(-7\)

    \[ -3 \times -4 = 12,\quad -3 + -4 = -7 \]

  2. Final answer

    \[ (x - 3)(x - 4) \]

Exam Traps

  • \(9x^2 + 15xy\) factorised fully is \(3x(3x + 5y)\), not \(x(9x + 15y)\). The number \(3\) is also a common factor.
  • When the quadratic starts \(2x^2 + \cdots\), factor \(2\) from the \(x\)-terms before you complete the square. Completing the square on \(2x^2 + 8x\) as if it were \(x^2 + 8x\) gives the wrong \(p\) and \(q\).

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