Algebra

• Two moduli: square. One modulus: split and check. A modulus inequality: sketch. A factor: \(p\!\left(-\tfrac{b}{a}\right) = 0\). A series: take out \(a^n\) and state where it is valid.
• Squaring is safe only when both sides are moduli. The remainder on dividing by \((ax + b)\) is \(p\!\left(-\tfrac{b}{a}\right)\). A repeated factor needs two fractions; a quadratic factor needs \(Bx + C\). Take out \(a^n\), then state \(|x| < \left|\tfrac{a}{b}\right|\).

• 3.1.1 2.1.1 understand the meaning of |x|, sketch the graph of y = |ax + b| and use relations such as |a| = |b| ⇔ a² = b² and |x − a| < b ⇔ a − b < x < a + b when solving equations and inequalities
• 3.1.2 2.1.2 divide a polynomial, of degree not exceeding 4, by a linear or quadratic polynomial, and identify the quotient and remainder (which may be zero)
• 3.1.3 2.1.3 use the factor theorem and the remainder theorem
• 3.1.4 Paper 3 only recall an appropriate form for expressing rational functions in partial fractions, and carry out the decomposition, in cases where the denominator is no more complicated than (ax + b)(cx + d)(ex + f), (ax + b)(cx + d)² or (ax + b)(cx² + d)
• 3.1.5 Paper 3 only use the expansion of (1 + x)ⁿ, where n is a rational number and |x| < 1

Accuracy for this chapter. Almost every answer here is exact: fractions such as \(\tfrac{4}{3}\), \(\tfrac{1}{64}\) and \(\tfrac{29}{16}\), not \(1.33\), \(0.0156\) or \(1.81\). Keep coefficients as fractions through the whole expansion. A decimal belongs only where a question asks for one (an estimate such as \(\sqrt{0.98}\) in the section E drill, to the accuracy it states) or in a check, where four significant figures are enough to show that the series and the function agree. Show the method line before every result: the squared equation, the cases, the substitution into \(p(x)\), the identity with its constants. A root found on a calculator's equation solver with no working earns nothing.

• “To solve \(2x + 5 < |x + 1|\), square both sides.” Correct Squaring an inequality is safe only when both sides are moduli (or are known to be non-negative). Here \(2x + 5\) can be negative, and squaring gives \(-4 < x < -2\), which loses every \(x \le -4\). Sketch the V and the line instead: the answer is \(x < -2\) (Figure 3).
• “The remainder when \(p(x)\) is divided by \((2x - 1)\) is \(p(1)\).” Correct It is \(p\!\left(\tfrac{1}{2}\right)\): substitute the value of \(x\) that makes the divisor zero. For \((ax + b)\) the remainder is \(p\!\left(-\tfrac{b}{a}\right)\).
• “\(|x| = \pm x\).” Correct \(|x|\) is one number, never negative: \(x\) if \(x \ge 0\), \(-x\) if \(x < 0\). The \(\pm\) belongs to the equation \(|x| = 3\), whose solutions are \(x = \pm 3\).
• “\(|2x - 1| = x + 4\) gives \(x = 5\) or \(x = -1\); done.” Correct With a modulus on one side only, each case must be checked in the original equation, because the right-hand side must not be negative. Here both pass; for \(|2x - 1| = x - 4\) both fail and there is no solution.
• “\(\dfrac{3x^2 + 3x + 3}{(x + 2)(x - 1)^2} = \dfrac{A}{x + 2} + \dfrac{B}{(x - 1)^2}\).” Correct A repeated factor needs both powers: \(\dfrac{B}{x - 1} + \dfrac{C}{(x - 1)^2}\). With only two constants the identity cannot be satisfied.
• “\((2 - \tfrac{1}{2}x)^{-1} = 1 - \tfrac{1}{2}x + \dots\)” Correct The MF19 series is for \((1 + x)^n\). Take out the 2 first: \(2^{-1}\left(1 - \tfrac{1}{4}x\right)^{-1}\), and then state the validity, \(|x| < 4\).
• “\(|x| = \pm x\).” Repair \(|x|\) is a single non-negative number: \(x\) if \(x \ge 0\), \(-x\) if \(x < 0\). The \(\pm\) belongs to the equation: \(|x| = 3 \Rightarrow x = \pm 3\).
• “\(|2x - 1| = x + 4 \Rightarrow x = 5\) or \(x = -1\); no check needed.” Repair The check is needed whenever one side has no modulus. Here both answers pass. But for \(|2x - 1| = x - 4\), the cases give \(x = -3\) and \(x = \tfrac{5}{3}\), and at both \(x - 4\) is negative, so both must be rejected: that equation has no solution.
• “\(2x + 5 < |x + 1|\); square: \((2x + 5)^2 < (x + 1)^2 \Rightarrow -4 < x < -2\).” Repair \(2x + 5\) can be negative, so squaring is not safe, and here it loses every \(x \le -4\). Sketch the V and the line; the answer is \(x < -2\).
• “\(|x - 3| < 2 \Rightarrow x < 5\).” Repair It is a two-sided condition, \(x\) within 2 of 3: \(1 < x < 5\).
• Leaving out the \(0x^2\) when dividing \(x^4 + 2x^3 - 3x + 7\). Repair Insert every missing power, or the columns misalign and the subtraction goes wrong: \(x^4 + 2x^3 + 0x^2 - 3x + 7\).
• “Dividing \(2x^4 - 3x^3 + x - 5\) by \(x^2 - 2\), the remainder is \(4x^2 - 5x - 5\).” Repair A remainder must have lower degree than the divisor. With a quadratic divisor you stop only when what is left is linear or constant; \(4x^2 - 5x - 5\) can be divided once more, giving \(+4\) in the quotient and the remainder \(-5x + 3\).
• “The remainder when \(p(x)\) is divided by \((2x - 1)\) is \(p(1)\).” Repair It is \(p\!\left(\tfrac{1}{2}\right)\), the value of \(x\) that makes the divisor zero.
• “\(p(2) = 0\), so \(x = 2\) is the answer” to “solve \(p(x) = 0\)”. Repair Divide out \((x - 2)\) and deal with the quotient. It may have more real roots, or you must show, with its discriminant, that it has none.
• Writing \(\dfrac{A}{x + 2} + \dfrac{B}{(x - 1)^2}\) for a repeated factor. Paper 3 only Repair The repeated factor needs two terms: \(\dfrac{B}{x - 1} + \dfrac{C}{(x - 1)^2}\).
• Writing \(\dfrac{A}{x + 1} + \dfrac{B}{x^2 + 4}\) for a quadratic factor. Paper 3 only Repair The numerator over a quadratic factor that does not factorise is linear: \(\dfrac{Bx + C}{x^2 + 4}\).
• “\((2 - \tfrac{1}{2}x)^{-1} = 1 - \tfrac{1}{2}x + \dots\)” Paper 3 only Repair Take out the 2 first: \(\tfrac{1}{2}\left(1 - \tfrac{1}{4}x\right)^{-1} = \tfrac{1}{2} + \tfrac{1}{8}x + \dots\).
• “\((1 + 2x)^{-\frac{1}{2}}\): the \(x^2\) term is \(\tfrac{3}{8}x^2\).” Paper 3 only Repair The \(x^2\) term is \(\dfrac{n(n - 1)}{2!}\) times \((2x)^2\), bracket included: \(\dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2} \times 4x^2 = \tfrac{3}{2}x^2\).
• An expansion with no validity statement, or “valid for \(|x| < 2\)” for a sum of pieces valid for \(|x| < 1\) and \(|x| < 2\). Paper 3 only Repair The series for \((1 + kx)^n\) is valid for \(|kx| < 1\), that is \(|x| < \tfrac{1}{|k|}\); write it every time. A sum is valid only where every piece is, so take the stricter condition: \(|x| < 1\).

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. When you reach chapter 13, re-answer worked example 6 and then integrate each piece. When you reach chapter 17, re-answer drill 6 of section C and find the two complex roots that the discriminant \(-16\) was hiding. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: the later Paper 3 chapters use these tools without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 3 content, examined in Paper 3 (and Paper 2 on the AS-only pure route). Paper 3 (Pure Mathematics 3) is compulsory for the A Level and is 30% of it. The Paper 2 (Pure Mathematics 2) outcomes this chapter also serves are marked in the syllabus map: Paper 2 is offered only in the AS-only Pure Mathematics route (Papers 1 and 2), where it is 40% of the AS Level, and that route cannot be carried forward to the A Level. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• Section 3.1 is a toolkit, and the syllabus says an individual question may involve ideas from more than one section, so its outcomes can be the parts of a longer structured question. The mixed challenge in this chapter is built that way: a show that \((x - a)\) is a factor followed by hence solve; a sketch of one or two V graphs followed by solving the equation or inequality they picture; two facts about factors or remainders turned into two equations for unknown coefficients; and, in Paper 3 only, partial fractions followed by hence an expansion and its range of validity. In each, the working to show is the method line: the substitution into \(p(x)\), the squared equation or the two cases, the identity with its constants, the adapted series.
• MF19 prints the series for \((1 + x)^n\) with \(n\) rational and \(|x| < 1\). It does not print the adaptation you need for \((a + bx)^n\): take out \(a^n\), use \(\tfrac{b}{a}x\) in place of \(x\), and state \(|x| < \left|\tfrac{a}{b}\right|\). The modulus results, the division identity, the factor and remainder theorems and the three partial-fraction forms must all be known; the MF19 card lists them.
• Almost every answer in this chapter is exact: fractions such as \(\tfrac{29}{16}\), surds such as \(\tfrac{1 \pm \sqrt{3}}{2}\), and inequalities with exact end points. Give a decimal only where a question asks for one, to 3 significant figures. The working that must be visible is the method line: the equation after squaring, each case and its check against the original, the value \(p\!\left(-\tfrac{b}{a}\right)\) written out, the constants of an identity, and the validity condition of every expansion. A root taken from a calculator's equation solver without that working earns nothing.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.