0580/21 May/June 2025 Worked Answers (IGCSE Maths 0580 Extended)

44 questions · 100 marks · 120 minutes

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1. Question 1

   2 marksAlgebraic simplification
   Step 1: Collect the $a$ terms: $9a + 3a = 12a$. Step 2: Collect the $b$ terms: $-7b + 2b = -5b$. Step 3: Combine: $12a - 5b$.
   Method:

   Identify like terms ($a$ terms and $b$ terms), add/subtract their coefficients separately, then write the simplified expression.
   Examiner tips

   • Always identify like terms before combining
   • Pay careful attention to signs when collecting terms

2. Question 2

   4 marksAngles in parallel lines
   Step 1: The angles on the lower parallel line are $65°$ and $140°$. The angle between the two transversals at the lower line is $180° - 65° - (180° - 140°) = 180° - 65° - 40° = 75°$. Step 2: Alternatively, the angle between the transversals at the lower intersection is $140° - 65° = 75°$. Step 3: Using co-interior angles or the triangle formed, the angle at the top is $180° - 65° - 40° = 75°$.
   Method:

   Use alternate angles and angles on a straight line to find the angle between the transversals at the point of intersection.
   Examiner tips

   • Label all angles on the diagram before calculating
   • Use alternate angles, co-interior angles, and angles on a straight line systematically

3. Question 3

   3 marksSimple interest
   Step 1: Calculate the simple interest: $I = \frac{P \times R \times T}{100} = \frac{2000 \times 4 \times 5}{100} = \$400$. Step 2: Add the interest to the principal: $2000 + 400 = \$2400$.
   Method:

   Apply the simple interest formula I = PRT/100, then add the interest to the principal to get the total value.
   Examiner tips

   • Read carefully whether the question asks for the interest or the total value
   • Simple interest means the same amount is earned each year

4. Question 4

   3 marksFractions and order of operations
   Step 1: Perform the multiplication first (BIDMAS): $\frac{3}{4} \times \frac{2}{5} = \frac{6}{20} = \frac{3}{10}$. Step 2: Now subtract: $\frac{7}{8} - \frac{3}{10}$. Find a common denominator: LCD = 40. Step 3: $\frac{7}{8} = \frac{35}{40}$ and $\frac{3}{10} = \frac{12}{40}$. Step 4: $\frac{35}{40} - \frac{12}{40} = \frac{23}{40}$.
   Method:

   Apply BIDMAS: multiply first, then subtract using a common denominator.
   Examiner tips

   • Always follow BIDMAS when operations are mixed
   • Simplify fractions after multiplication before subtracting

5. Question 5

   2 marksInterior angles of regular polygons
   Step 1: The exterior angle is $180° - 160° = 20°$. Step 2: The number of sides is $\frac{360°}{20°} = 18$.
   Method:

   Calculate the exterior angle by subtracting the interior angle from 180, then divide 360 by the exterior angle.
   Examiner tips

   • Remember: exterior angle = 180 minus interior angle
   • Sum of exterior angles is always 360 degrees for any convex polygon

6. Question 6a

   2 marksDrawing straight line graphs
   Step 1: Substitute $x = 2$: $y = 3(2) - 2 = 6 - 2 = 4$. So $(2, 4)$ is on the line. Step 2: Check $(1, 3)$: $y = 3(1) - 2 = 1 \neq 3$. Not on the line. Step 3: Check $(0, 2)$: $y = 3(0) - 2 = -2 \neq 2$. Not on the line. Step 4: Check $(3, 6)$: $y = 3(3) - 2 = 7 \neq 6$. Not on the line.
   Method:

   Substitute the x-coordinate of each point into y = 3x - 2 and check whether the result matches the y-coordinate.
   Examiner tips

   • To check if a point lies on a line, substitute the x-value and see if you get the y-value
   • Plot at least two points to draw a straight line

7. Question 6b

   1 marksSimultaneous equations (graphical)
   Step 1: Set the equations equal: $x + 3 = -2x + 9$. Step 2: Solve: $3x = 6$, so $x = 2$. Step 3: Substitute: $y = 2 + 3 = 5$. Step 4: Therefore $a + b = 2 + 5 = 7$.
   Method:

   Set the two y-expressions equal, solve for x, then find y and compute a + b.
   Examiner tips

   • The intersection point of two lines gives the solution to the simultaneous equations
   • Always check your solution by substituting back into both equations

8. Question 7

   3 marksRecurring decimals to fractions
   Step 1: Let $x = 0.18888...$ Step 2: $10x = 1.8888...$ Step 3: $100x = 18.8888...$ Step 4: Subtract: $100x - 10x = 18.888... - 1.888... = 17$ Step 5: $90x = 17$, so $x = \frac{17}{90}$.
   Method:

   Let x = the decimal, multiply by powers of 10 to align recurring parts, subtract to eliminate the recurring portion, then simplify.
   Examiner tips

   • Identify which digits recur and which do not
   • Multiply by the appropriate power of 10 to shift the decimal point past the non-recurring part

9. Question 8a

   2 marksVector arithmetic
   Step 1: Calculate $2\mathbf{a} = 2\begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 10 \\ 6 \end{pmatrix}$. Step 2: Subtract $\mathbf{b}$: $\begin{pmatrix} 10 \\ 6 \end{pmatrix} - \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 10 - (-2) \\ 6 - 1 \end{pmatrix} = \begin{pmatrix} 12 \\ 5 \end{pmatrix}$.
   Method:

   Multiply vector a by 2, then subtract vector b component-wise, being careful with signs.
   Examiner tips

   • When subtracting a negative component, it becomes addition
   • Write out each step to avoid sign errors

10. Question 8b

    2 marksMagnitude of a vector
    Step 1: Use the magnitude formula: $\sqrt{3^2 + p^2} = 5$. Step 2: Square both sides: $9 + p^2 = 25$. Step 3: Solve: $p^2 = 16$, so $p = 4$ (since $p > 0$).
    Method:

    Set up the equation using the magnitude formula, square both sides, solve for the unknown, and take the positive square root.
    Examiner tips

    • The magnitude formula is essentially Pythagoras' theorem applied to vectors
    • Remember to take the square root at the end

11. Question 9

    3 marksMean from frequency table
    Step 1: Total frequency: $4 + k + 6 + 3 = 13 + k = 20$, so $k = 7$. Step 2: Verify with mean: $\frac{0(4) + 1(7) + 2(6) + 3(3)}{20} = \frac{0 + 7 + 12 + 9}{20} = \frac{28}{20} = 1.4$. Note: Using the total frequency condition gives $k = 7$ directly. The mean can be used as a check.
    Method:

    Use the mean formula: mean = sum(fx)/sum(f). Set up the equation and solve for k.
    Examiner tips

    • Mean = sum of (value x frequency) / total frequency
    • Check your answer by verifying the mean with the calculated value

12. Question 10a

    1 marksCircle theorems - angle at centre
    Step 1: The angle at the centre is twice the angle at the circumference subtended by the same arc. Step 2: Angle at centre $= 2 \times 35° = 70°$.
    Method:

    Apply the circle theorem: angle at centre = 2 x angle at circumference.
    Examiner tips

    • Always identify which arc the angles are subtended by
    • State the circle theorem you are using

13. Question 10b

    1 marksCircle theorems - angle in semicircle
    Step 1: The angle at the circumference is half the angle at the centre. Step 2: Angle at circumference $= \frac{110°}{2} = 55°$.
    Method:

    Apply the theorem: angle at circumference = half the angle at the centre for the same arc.
    Examiner tips

    • Identify whether the angle at the centre is a reflex angle or not
    • The angle at the circumference is half the angle at the centre for the same arc

14. Question 10c

    1 marksCircle theorems - tangent-chord angle
    Step 1: By the alternate segment theorem, the angle between a tangent and a chord equals the angle in the alternate segment. Step 2: Therefore the angle between the tangent and chord is $55°$.
    Method:

    Apply the alternate segment theorem: angle between tangent and chord = angle in alternate segment.
    Examiner tips

    • The alternate segment theorem is frequently tested
    • A tangent is always perpendicular to the radius at the point of contact

15. Question 10d

    1 marksCircle theorems
    Step 1: The difference between the two angles is $55° - 40° = 15°$.
    Method:

    Use the angles found in previous parts to calculate the required angle by subtraction.
    Examiner tips

    • In multi-part circle theorem questions, use your answers from earlier parts
    • Clearly label all angles on the diagram

16. Question 11a

    2 marksIdentifying points on a curve
    Step 1: Substitute $y = 5$: $5 = a^2 + 1$. Step 2: Solve: $a^2 = 4$, so $a = 2$ (since $a > 0$).
    Method:

    Substitute the known coordinate into the curve equation and solve for the unknown.
    Examiner tips

    • Read coordinates carefully from graphs
    • Check your answer by substituting back into the equation

17. Question 11b

    3 marksEquation of a tangent
    Step 1: Use $y - y_1 = m(x - x_1)$ with $m = -4$ and $(x_1, y_1) = (2, 3)$. Step 2: $y - 3 = -4(x - 2)$. Step 3: $y - 3 = -4x + 8$. Step 4: $y = -4x + 11$.
    Method:

    Substitute the gradient and point into y - y1 = m(x - x1), then rearrange to y = mx + c form.
    Examiner tips

    • Always use y - y1 = m(x - x1) when given a point and gradient
    • Be careful with signs when expanding brackets

18. Question 12a

    2 marksCumulative frequency - interquartile range
    Step 1: Interquartile range = Upper quartile $-$ Lower quartile. Step 2: IQR $= 28 - 12 = 16$ minutes.
    Method:

    Read the lower quartile and upper quartile from the cumulative frequency curve, then subtract to find the IQR.
    Examiner tips

    • Find Q1 at 1/4 of the total frequency and Q3 at 3/4 of the total frequency
    • Draw lines on the graph to show your readings clearly

19. Question 12b

    2 marksCumulative frequency - percentiles
    Step 1: Parcels with mass $\geq m$: $60 - 45 = 15$. Step 2: Fraction $= \frac{15}{60} = \frac{1}{4}$.
    Method:

    Read the cumulative frequency at the given value, subtract from total to find the number above, then express as a fraction of the total.
    Examiner tips

    • Always use the total frequency as the denominator
    • Read carefully whether the question asks for 'less than' or 'greater than'

20. Question 13a

    5 marksSimilar shapes - frustum
    Step 1: The cones are similar, so the ratio of corresponding lengths is constant. Step 2: Scale factor $= \frac{10}{4} = \frac{5}{2}$. Step 3: Height of larger cone $= 6 \times \frac{5}{2} = 15$ cm.
    Method:

    Find the linear scale factor from the given radii, then multiply the known height by this factor.
    Examiner tips

    • Always identify the scale factor first in similar shapes problems
    • Use the linear scale factor for lengths, squared for areas, cubed for volumes

21. Question 13b

    3 marksPythagoras' theorem in 3D
    Step 1: The radius, height, and slant height form a right triangle. Step 2: $l^2 = r^2 + h^2 = 6^2 + 8^2 = 36 + 64 = 100$. Step 3: $l = \sqrt{100} = 10$ cm.
    Method:

    Identify the right triangle formed by the radius, height, and slant height. Apply Pythagoras' theorem to find the slant height.
    Examiner tips

    • Draw the right triangle within the cone to visualize the problem
    • Check whether you need to add or subtract the squares

22. Question 14a

    2 marksFunction evaluation
    Step 1: Substitute $x = -1$: $f(-1) = 3(-1)^2 - 7$. Step 2: $(-1)^2 = 1$, so $f(-1) = 3(1) - 7 = 3 - 7 = -4$.
    Method:

    Substitute x = -1 into the function, being careful with the sign when squaring.
    Examiner tips

    • Always use brackets when substituting negative values
    • Remember: (-1)^2 = 1, not -1

23. Question 14b

    2 marksInverse functions
    Step 1: Let $y = 5x - 2$. Step 2: Swap $x$ and $y$: $x = 5y - 2$. Step 3: Solve for $y$: $x + 2 = 5y$, so $y = \frac{x + 2}{5}$. Step 4: Therefore $g^{-1}(x) = \frac{x + 2}{5}$.
    Method:

    Replace f(x) with y, swap x and y, then rearrange to isolate y.
    Examiner tips

    • Write y = f(x), swap x and y, then solve for y
    • Check by verifying f(f^{-1}(x)) = x

24. Question 14c

    2 marksComposite functions
    Step 1: $fg(x) = f(g(x)) = f(3x - 2)$. Step 2: Substitute into $f$: $2(3x - 2) + 1 = 6x - 4 + 1 = 6x - 3$. Step 3: So $a = 6$ and $b = -3$.
    Method:

    Substitute g(x) into f(x), expand, and simplify to find the coefficients a and b.
    Examiner tips

    • fg(x) means f(g(x)) - apply g first, then f
    • Expand brackets carefully and collect like terms

25. Question 14d

    3 marksAlgebraic fractions
    Step 1: Common denominator is $(2x + 1)(x - 3)$. Step 2: $\frac{3(x - 3) - 2(2x + 1)}{(2x + 1)(x - 3)}$. Step 3: Expand numerator: $3x - 9 - 4x - 2 = -x - 11$. Step 4: Result: $\frac{-x - 11}{(2x + 1)(x - 3)}$.
    Method:

    Find the LCD, rewrite each fraction with the common denominator, expand the numerators, and simplify.
    Examiner tips

    • Always expand brackets fully before simplifying
    • Be very careful with signs when subtracting fractions

26. Question 15a

    2 marksSurds - expansion
    Step 1: Expand using FOIL: $(4)(3) + (4)(\sqrt{3}) + (-\sqrt{3})(3) + (-\sqrt{3})(\sqrt{3})$. Step 2: $= 12 + 4\sqrt{3} - 3\sqrt{3} - 3$. Step 3: $= (12 - 3) + (4\sqrt{3} - 3\sqrt{3}) = 9 + \sqrt{3}$.
    Method:

    Expand using FOIL, then collect rational terms and surd terms separately.
    Examiner tips

    • Write out all four terms from the expansion before simplifying
    • Remember sqrt(a) x sqrt(a) = a, not a^2

27. Question 15b

    2 marksRationalising the denominator
    Step 1: Multiply by $\frac{\sqrt{6}}{\sqrt{6}}$: $\frac{4}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{4\sqrt{6}}{6}$. Step 2: Simplify: $\frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3}$.
    Method:

    Multiply numerator and denominator by the surd, then simplify the fraction.
    Examiner tips

    • Always simplify the fraction after rationalising
    • Rationalising means removing the surd from the denominator

28. Question 16

    3 marksExpanding triple brackets
    Step 1: First expand $(x + 2)(x - 1) = x^2 + x - 2$. Step 2: Now multiply by $(2x + 3)$: $(x^2 + x - 2)(2x + 3)$. Step 3: $= 2x^3 + 3x^2 + 2x^2 + 3x - 4x - 6$. Step 4: $= 2x^3 + 5x^2 - x - 6$.
    Method:

    Expand two of the three brackets first, simplify, then multiply the result by the remaining bracket.
    Examiner tips

    • Expand two brackets first, simplify, then multiply by the third
    • Check your answer by substituting a value, e.g. x = 1

29. Question 17a

    2 marksProbability
    Step 1: $P(\text{red first}) = \frac{3}{10}$. Step 2: With replacement, $P(\text{red second}) = \frac{3}{10}$. Step 3: $P(\text{both red}) = \frac{3}{10} \times \frac{3}{10} = \frac{9}{100}$.
    Method:

    Find the probability of each event separately, then multiply since the events are independent (with replacement).
    Examiner tips

    • Check whether the question says 'with replacement' or 'without replacement'
    • For independent events, multiply the probabilities

30. Question 17b_i

    2 marksProbability - tree diagrams
    Step 1: $P(\text{no rain Monday}) = 1 - P(\text{rain Monday}) = 1 - \frac{3}{5} = \frac{2}{5}$.
    Method:

    Use the fact that P(event) + P(not event) = 1 to find the complementary probability.
    Examiner tips

    • Probabilities on each pair of branches must sum to 1
    • Read carefully which probabilities are conditional

31. Question 17b_ii

    3 marksProbability - combined events
    Step 1: $P(\text{pass 1st}) = \frac{2}{3}$. Step 2: $P(\text{fail 1st then pass 2nd}) = \frac{1}{3} \times \frac{3}{4} = \frac{3}{12} = \frac{1}{4}$. Step 3: $P(\text{pass within 2}) = \frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}$.
    Method:

    Identify all paths leading to success, multiply probabilities along each path, then add the results.
    Examiner tips

    • List all the paths that lead to the desired outcome
    • Multiply along branches (AND), add between paths (OR)

32. Question 18a

    1 marksForming expressions
    Step 1: Time $= \frac{\text{Distance}}{\text{Speed}} = \frac{80}{v}$.
    Method:

    Apply the formula Time = Distance / Speed with the given values.
    Examiner tips

    • Remember the speed-distance-time triangle
    • Time = Distance / Speed

33. Question 18b

    1 marksForming expressions
    Step 1: Time $= \frac{\text{Distance}}{\text{Speed}} = \frac{60}{s + 5}$.
    Method:

    Apply Time = Distance / Speed, keeping the compound speed expression together.
    Examiner tips

    • Keep the algebraic expression for speed together in brackets in the denominator
    • Do not separate terms when they form a single expression

34. Question 18c

    4 marksForming and solving equations
    Step 1: Set up the equation: $\frac{10}{x-4} - \frac{12}{x} = 1$. Step 2: Multiply through by $x(x-4)$: $10x - 12(x-4) = x(x-4)$. Step 3: Expand: $10x - 12x + 48 = x^2 - 4x$. Step 4: Simplify: $-2x + 48 = x^2 - 4x$, giving $x^2 - 2x - 48 = 0$.
    Method:

    Set up the equation using the time expressions, multiply through by the LCD, expand, and rearrange into standard quadratic form.
    Examiner tips

    • In 'show that' questions, you must show every step of working
    • Multiply both sides by the LCM of the denominators to clear fractions

35. Question 18d

    3 marksSolving quadratic equations by factorisation
    Step 1: Find two numbers that multiply to $-28$ and add to $+3$: $7$ and $-4$. Step 2: Factorise: $(x + 7)(x - 4) = 0$. Step 3: $x + 7 = 0$ gives $x = -7$. $x - 4 = 0$ gives $x = 4$.
    Method:

    Find two numbers with the required product and sum, write the factorised form, then solve each bracket equal to zero.
    Examiner tips

    • Check your factors: they must multiply to give the constant term and add to give the x coefficient
    • Don't forget to write both solutions

36. Question 18e

    1 marksInterpreting solutions in context
    Step 1: Since speed must be positive, $x = 8$ km/h. Step 2: Time $= \frac{12}{8} = 1.5$ hours.
    Method:

    Reject the negative solution, then use the positive value to calculate the time.
    Examiner tips

    • Always consider the context when choosing between solutions
    • Reject solutions that don't make physical sense (negative speeds, times, lengths)

37. Question 19

    2 marksIndices and powers
    Step 1: $27^{-\frac{2}{3}} = \frac{1}{27^{\frac{2}{3}}}$. Step 2: $27^{\frac{1}{3}} = \sqrt[3]{27} = 3$. Step 3: $27^{\frac{2}{3}} = 3^2 = 9$. Step 4: $27^{-\frac{2}{3}} = \frac{1}{9}$.
    Method:

    Deal with the negative sign (reciprocal), find the cube root, then square the result.
    Examiner tips

    • Negative index means reciprocal, fractional index means root and power
    • Find the root first, then raise to the power, then take the reciprocal

38. Question 20

    4 marksTrigonometry - exact values
    Step 1: Using $\cos 30° = \frac{\text{adjacent}}{\text{hypotenuse}}$. Step 2: $\cos 30° = \frac{\sqrt{3}}{2}$, so $\frac{\sqrt{3}}{2} = \frac{9}{h}$. Step 3: $h = \frac{9 \times 2}{\sqrt{3}} = \frac{18}{\sqrt{3}} = \frac{18\sqrt{3}}{3} = 6\sqrt{3}$ cm.
    Method:

    Identify the correct trig ratio, use the exact value of \cos 30, rearrange for the hypotenuse, and rationalise.
    Examiner tips

    • Know the exact values for sin, cos, tan of 30, 45, and 60 degrees
    • Rationalise the denominator when giving exact answers with surds

39. Question 21a_i

    1 marksVectors - path finding
    Step 1: $\overrightarrow{BA} = \overrightarrow{BO} + \overrightarrow{OA}$. Step 2: $\overrightarrow{BO} = -\overrightarrow{OB} = -\mathbf{b}$. Step 3: $\overrightarrow{BA} = -\mathbf{b} + \mathbf{a} = \mathbf{a} - \mathbf{b}$.
    Method:

    Find a path from B to A via O, using known vectors and reversing directions as needed.
    Examiner tips

    • Use the vector path: go via a known point (usually O)
    • Reversing a vector changes its sign

40. Question 21a_ii

    1 marksVectors - path finding
    Step 1: $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM}$. Step 2: $\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB} = \frac{1}{2}(\mathbf{b} - \mathbf{a})$. Step 3: $\overrightarrow{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \mathbf{a} + \frac{1}{2}\mathbf{b} - \frac{1}{2}\mathbf{a} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$.
    Method:

    Find the path O to A to M, where M is the midpoint of AB.
    Examiner tips

    • The midpoint formula for vectors: M = 1/2(A + B) when expressed from the origin
    • Always check your path makes geometric sense

41. Question 21b

    3 marksVectors - proving geometric properties
    Step 1: $\overrightarrow{SR} = 4\mathbf{a} = 2 \times 2\mathbf{a} = 2\overrightarrow{PQ}$. Step 2: Since $\overrightarrow{SR}$ is a scalar multiple of $\overrightarrow{PQ}$, they are parallel. Step 3: $PQ \neq SR$ (different lengths), so exactly one pair of sides is parallel, making $PQRS$ a trapezium.
    Method:

    Find the vectors for opposite sides, show one is a scalar multiple of the other (parallel), then conclude it is a trapezium.
    Examiner tips

    • To prove parallel: show one vector is a scalar multiple of the other
    • State clearly which sides are parallel and why this makes it a trapezium

42. Question 22a

    2 marksTurning points of quadratics
    Step 1: The completed square form $y = (x - p)^2 + q$ has its turning point at $(p, q)$. Step 2: Since the turning point is $(4, -5)$, we have $p = 4$ and $q = -5$.
    Method:

    Identify that the completed square form y = (x - p)^2 + q has turning point (p, q) and read off the values.
    Examiner tips

    • The turning point of y = (x - p)^2 + q is (p, q)
    • Be careful with the sign of p: (x - 4)^2 means p = 4, not -4

43. Question 22b

    4 marksSolving quadratic inequalities / finding intersections
    Step 1: Set the curve equal to the line: $(x - 3)^2 = -2x + 6$. Step 2: Expand: $x^2 - 6x + 9 = -2x + 6$. Step 3: Rearrange: $x^2 - 4x + 3 = 0$. Step 4: Factorise: $(x - 1)(x - 3) = 0$, so $x = 1$ or $x = 3$. Step 5: When $x = 1$: $y = -2(1) + 6 = 4$. When $x = 3$: $y = -2(3) + 6 = 0$. Step 6: Intersection points: $(1, 4)$ and $(3, 0)$.
    Method:

    Set the curve and line equal, form and solve the resulting quadratic, then find the y-values by substituting back.
    Examiner tips

    • Always substitute back to find both coordinates
    • Check your answers satisfy both equations

44. Question 23

    3 marksSimplifying algebraic fractions
    Step 1: Factorise the numerator: $3x^2 + 6x = 3x(x + 2)$. Step 2: Factorise the denominator: $3x^2 - 12 = 3(x^2 - 4) = 3(x + 2)(x - 2)$. Step 3: Cancel common factors: $\frac{3x(x + 2)}{3(x + 2)(x - 2)} = \frac{x}{x - 2}$.
    Method:

    Factorise both numerator and denominator completely, identify common factors, then cancel.
    Examiner tips

    • Always factorise fully before cancelling
    • Look for common factors, difference of two squares, and quadratic factorisations
    • Never cancel individual terms - only cancel factors