Algebra

Syllabus coverage

Understand the meaning of $| x |$ , sketch the graph of $y = | a x + b |$ , and use $| a | = | b | \Leftrightarrow a^{2} = b^{2}$ and $| x - a | < b \Leftrightarrow a - b < x < a + b$ when solving equations and inequalities
Divide a polynomial of degree not exceeding 4 by a linear or a quadratic polynomial, and identify the quotient and the remainder, which may be zero
Use the factor theorem and the remainder theorem to find factors and remainders, solve polynomial equations and evaluate unknown coefficients
Recall an appropriate form for expressing rational functions in partial fractions and carry out the decomposition, for denominators of the form $(a x + b) (c x + d) (e x + f)$ , $(a x + b) (c x + d)^{2}$ and $(a x + b) (c x^{2} + d)$
Use the expansion of $(1 + x)^{n}$ , where $n$ is a rational number and $| x | < 1$ , including adapting the standard series and stating the values of $x$ for which the expansion is valid

Part of the AS & A Level Maths syllabus 9709.

The graph of

y = | 2 x - 3 |

is the line

y = 2 x - 3

with the part below the

x

-axis reflected upwards. The vertex is at

(1.5, 0)

and the

y

-intercept at

(0, 3)

The modulus function and the graph of y = |ax + b|

The modulus of a number is its distance from zero, so it is never negative. $| 7 | = 7$ and $| - 7 | = 7$ . Written as a definition that a mark scheme will accept: $| x | = x$ when $x \geq 0$ , and $| x | = - x$ when $x < 0$ . The second line is the one students misread. For $x = - 7$ the rule gives $| - 7 | = - (- 7) = 7$ , not $- 7$ ; the minus sign in the definition is doing the work of removing the sign of the number inside.

The graph of $y = | a x + b |$ is built from the graph of the straight line $y = a x + b$ . Draw the line, then reflect in the x-axis every part of it that lies below the axis. The result is a V shape whose lowest point - the vertex - sits on the x-axis at the value where the inside of the modulus is zero, that is at $x = - \frac{b}{a}$ . For $y = | 2 x - 3 |$ that is $x = \frac{3}{2}$ , so the vertex is $(\frac{3}{2}, 0)$ .

Two more facts fix the sketch. Putting $x = 0$ gives height $| b |$ , so $y = | 2 x - 3 |$ meets the y-axis at $(0, 3)$ . Either side of the vertex the graph is straight: on the right it is the line $y = a x + b$ with gradient $a$ , and on the left it is the reflection with gradient $- a$ . Cambridge's note for this section is explicit that graphs of $y = | f (x) |$ and $y = f (| x |)$ for non-linear $f$ are not included, so every sketch you are asked for is this same V shape.

A sketch is marked on the features, not on the graph paper. Label the vertex, label the intercepts, and make it obvious that both arms are straight lines meeting at a point on the x-axis. A curve that is smoothly rounded at the bottom, or a V that does not touch the x-axis, is a wrong answer however tidy it looks.

Feature of the graph	How to find it	For $y = \| 2 x - 3 \|$
Vertex (the corner)	solve $a x + b = 0$	$(\frac{3}{2}, 0)$
y-intercept	put $x = 0$ ; the height is $\| b \|$	$(0, 3)$
Gradient to the right of the vertex	$+ a$	$2$
Gradient to the left of the vertex	$- a$	$- 2$
Range	$y \geq 0$ always	$y \geq 0$

A modulus is never negative, so an equation such as

| 3 x - 2 | = - 5

has no solutions at all. Noticing that in the first line of a question saves five minutes of algebra that was never going to work.

Solving equations and inequalities with modulus

Two standard results do nearly all the work. First, $| a | = | b | \Leftrightarrow a^{2} = b^{2}$ , because squaring destroys the sign. Second, when $b > 0$ , $| x - a | < b \Leftrightarrow a - b < x < a + b$ , and $| x - a | > b \Leftrightarrow x < a - b$ or $x > a + b$ . The syllabus prints both in this form and expects you to use them rather than to redraw a graph for every question.

For $| 3 x - 2 | = | 2 x + 7 |$ the squaring result is the quickest route: $(3 x - 2)^{2} = (2 x + 7)^{2}$ , so $9 x^{2} - 12 x + 4 = 4 x^{2} + 28 x + 49$ , which gives $5 x^{2} - 40 x - 45 = 0$ and then $x^{2} - 8 x - 9 = 0$ , that is $(x - 9) (x + 1) = 0$ . So $x = 9$ or $x = - 1$ . Both check: at $x = 9$ the two sides of the original equation are 25 and 25, and at $x = - 1$ they are 5 and 5.

Inequalities need one extra thought. Squaring both sides is safe only when both sides are certainly non-negative, and in a question such as $2 x + 5 < | x + 1 |$ the left-hand side is negative once $x < - 2.5$ . Split the problem at the value where the inside of the modulus changes sign, which is $x = - 1$ . If $x \geq - 1$ then $| x + 1 | = x + 1$ , so $2 x + 5 < x + 1$ gives $x < - 4$ , contradicting $x \geq - 1$ , so this branch contributes nothing. If $x < - 1$ then $| x + 1 | = - (x + 1)$ , so $2 x + 5 < - x - 1$ , giving $3 x < - 6$ , that is $x < - 2$ . The full solution is $x < - 2$ .

Always test a value from your answer and a value from outside it. Take $x = - 3$ : the left side is $- 1$ and the right is $2$ , and $- 1 < 2$ is true, so $- 3$ belongs. Take $x = 0$ , which you are claiming is excluded: $5 < 1$ is false. The boundary $x = - 2$ is itself excluded because the inequality is strict, so the answer is written $x < - 2$ and never $x \leq - 2$ .

Statement	Equivalent form	When it applies
$\| a \| = \| b \|$	$a^{2} = b^{2}$	any values of $a$ and $b$
$\| x - a \| < b$	$a - b < x < a + b$	$b > 0$
$\| x - a \| > b$	$x < a - b$ or $x > a + b$	$b > 0$
$\| u \| = k$	$u = k$ or $u = - k$	$k > 0$ ; no solutions if $k < 0$
$\| u \| < k$	$- k < u < k$	$k > 0$

Never square an inequality unless you are certain both sides are non-negative.

2 x + 5 < | x + 1 |

is true at

x = - 5

, yet squaring both sides there would give

25 < 16

, which is false. Split into cases instead.

Dividing polynomials: quotient and remainder

Polynomial long division is the same process as dividing numbers, and the layout is worth copying from the mark scheme: write the dividend and the divisor, build the quotient one term at a time, and subtract at every stage. The identity underneath it is $P (x) = (divisor) \times (quotient) + remainder$ . Paper 3 divides a polynomial of degree at most 4 either by a linear polynomial such as $x - 3$ or $2 x + 1$ , or by a quadratic such as $x^{2} + 1$ .

Worked example: divide $2 x^{3} + x^{2} - 4 x + 5$ by $x + 1$ . The leading term is $2 x^{3} \div x = 2 x^{2}$ , and $2 x^{2} (x + 1) = 2 x^{3} + 2 x^{2}$ ; subtracting leaves $- x^{2} - 4 x$ . Next $- x^{2} \div x = - x$ , and $- x (x + 1) = - x^{2} - x$ ; subtracting leaves $- 3 x + 5$ . Next $- 3 x \div x = - 3$ , and $- 3 (x + 1) = - 3 x - 3$ ; subtracting leaves 8, a constant, which cannot be divided by $x$ again. The quotient is $2 x^{2} - x - 3$ and the remainder is 8.

Check by multiplying back: $(x + 1) (2 x^{2} - x - 3) + 8 = 2 x^{3} - x^{2} - 3 x + 2 x^{2} - x - 3 + 8 = 2 x^{3} + x^{2} - 4 x + 5$ . Thirty seconds of checking is worth more than running the whole division a second time.

Dividing by a quadratic follows the same steps with a longer first product. Divide $x^{4} - 3 x^{2} + 2 x - 1$ by $x^{2} + 1$ : $x^{4} \div x^{2} = x^{2}$ , and $x^{2} (x^{2} + 1) = x^{4} + x^{2}$ ; subtracting leaves $- 4 x^{2} + 2 x - 1$ . Then $- 4 x^{2} \div x^{2} = - 4$ , and $- 4 (x^{2} + 1) = - 4 x^{2} - 4$ ; subtracting leaves $2 x + 3$ . The quotient is $x^{2} - 4$ and the remainder is $2 x + 3$ . Notice the shape of the answer: a quadratic divisor leaves a remainder of degree at most 1, so the remainder is linear, not a single number.

Dividing by $x - k$ leaves a constant remainder, and that constant is $f (k)$
Dividing by $x^{2} + p x + q$ leaves a remainder of the form $r x + s$
The degree of the quotient is the degree of the dividend minus the degree of the divisor
If the remainder is zero, the divisor is a factor of the polynomial - say so in words, because the mark scheme looks for it
A missing power must still be written in: divide $x^{4} - 3 x^{2} + 2 x - 1$ with the $x^{3}$ column kept as $0 x^{3}$ so the columns line up

Write the quotient and the remainder as two separate statements. 'Quotient

2 x^{2} - x - 3

, remainder 8' is unambiguous; a single line of algebra with a stray

+ 8

tacked on the end often loses the remainder mark.

The remainder theorem and the factor theorem

The remainder theorem is the shortcut for that constant remainder: when $f (x)$ is divided by $x - k$ , the remainder is $f (k)$ . It follows in one line from the division identity $f (x) = (x - k) q (x) + r$ by putting $x = k$ , because the first term vanishes and $f (k) = r$ . Dividing by $a x - b$ rather than $x - k$ gives the remainder $f (\frac{b}{a})$ : set the divisor to zero and solve, which gives the value to substitute.

The factor theorem is the same statement read the other way round. $x - k$ is a factor of $f (x)$ exactly when $f (k) = 0$ , and $a x - b$ is a factor exactly when $f (\frac{b}{a}) = 0$ . A remainder of zero and a factor are the same piece of information, which is why one identity answers both questions.

Solving a cubic is a three-step routine. First find one root by trying the factors of the constant term, divided by the factors of the leading coefficient if that coefficient is not 1. Then confirm it with $f (k) = 0$ and divide by $x - k$ to reduce the cubic to a quadratic. Finally factorise the quadratic or use the quadratic formula. For $f (x) = x^{3} - 3 x^{2} - 4 x + 12$ , $f (2) = 8 - 12 - 8 + 12 = 0$ , and dividing by $x - 2$ gives $x^{2} - x - 6 = (x - 3) (x + 2)$ , so the roots are 2, 3 and $- 2$ .

Unknown coefficients come from writing one equation for every fact you are given. If $f (x) = x^{3} + a x + b$ has $x - 1$ as a factor and leaves a remainder of 10 when divided by $x - 2$ , then $f (1) = 1 + a + b = 0$ and $f (2) = 8 + 2 a + b = 10$ . Subtracting the first from the second gives $a = 3$ , and then $b = - 4$ . Two facts, two equations, two unknowns - and the equations are always linear.

Integer candidates for a root are the factors of the constant term, positive and negative
If the leading coefficient is not 1, also try fractions such as $\frac{1}{2}$ , $\frac{3}{2}$
Always show the substitution: 'M1 for attempting $f (2)$ ' is a mark you can only earn on paper
'Given that $x - 2$ is a factor' means $f (2) = 0$ ; the trial has been done for you
Once you have one factor, divide rather than guess the other two

A cubic with a repeated root still has three roots, and a cubic with no rational root still has a root that the factor theorem cannot find. On Paper 3 every cubic in this section factors into a linear and a quadratic with rational roots.

Partial fractions: the three denominators Paper 3 uses

A rational function is a fraction with polynomials above and below the line. Expressing it in partial fractions means writing it as a sum of simpler fractions whose denominators are the factors of the original denominator. The syllabus asks you to recall an appropriate form, so the form of the answer is itself worth a mark: the number of constants you need equals the degree of the denominator.

The method never changes. Multiply both sides by the denominator, which turns the identity into a polynomial identity with no fractions in it, then either substitute the values that make each factor zero - the cover-up shortcut - or compare coefficients. Substituting is faster and safer; comparing coefficients is the fallback when the substitutions are awkward.

Worked example with a coefficient of $x$ that is not 1: $\frac{4 x + 1}{(2 x + 1) (x - 1)} = \frac{A}{2 x + 1} + \frac{B}{x - 1}$ , so $4 x + 1 = A (x - 1) + B (2 x + 1)$ . Put $x = 1$ : $5 = 3 B$ , so $B = \frac{5}{3}$ . Put $x = - \frac{1}{2}$ : $- 1 = - \frac{3}{2} A$ , so $A = \frac{2}{3}$ . The decomposition is $\frac{2}{3 (2 x + 1)} + \frac{5}{3 (x - 1)}$ .

Worked example with a repeated factor: $\frac{3 x^{2} + 3 x - 9}{(x + 1)^{2} (x - 2)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^{2}} + \frac{C}{x - 2}$ , so $3 x^{2} + 3 x - 9 = A (x + 1) (x - 2) + B (x - 2) + C (x + 1)^{2}$ . Put $x = - 1$ : $3 - 3 - 9 = - 9 = - 3 B$ , so $B = 3$ . Put $x = 2$ : $12 + 6 - 9 = 9 = 9 C$ , so $C = 1$ . With $B$ and $C$ known, put $x = 0$ : $- 9 = - 2 A - 6 + 1$ , so $A = 2$ . The decomposition is $\frac{2}{x + 1} + \frac{3}{(x + 1)^{2}} + \frac{1}{x - 2}$ .

Worked example with a quadratic factor: $\frac{3 x^{2} + 5 x + 7}{(x + 1) (x^{2} + 4)} = \frac{A}{x + 1} + \frac{B x + C}{x^{2} + 4}$ , so $3 x^{2} + 5 x + 7 = A (x^{2} + 4) + (B x + C) (x + 1)$ . Put $x = - 1$ : $5 = 5 A$ , so $A = 1$ . Compare the $x^{2}$ coefficients: $3 = A + B$ , so $B = 2$ . Compare the constant terms: $7 = 4 A + C$ , so $C = 3$ . The decomposition is $\frac{1}{x + 1} + \frac{2 x + 3}{x^{2} + 4}$ .

One exclusion is worth knowing. Cases where the degree of the numerator is greater than or equal to the degree of the denominator are not examined at 9709, so the fraction you are handed is always proper. If it ever is not, dividing first would be the fix - but you will not be asked for it here.

Denominator	The form to write down	Constants needed
$(a x + b) (c x + d)$	$\frac{A}{a x + b} + \frac{B}{c x + d}$	2
$(a x + b) (c x + d) (e x + f)$	$\frac{A}{a x + b} + \frac{B}{c x + d} + \frac{C}{e x + f}$	3
$(a x + b) (c x + d)^{2}$	$\frac{A}{a x + b} + \frac{B}{c x + d} + \frac{C}{(c x + d)^{2}}$	3
$(a x + b) (c x^{2} + d)$	$\frac{A}{a x + b} + \frac{B x + C}{c x^{2} + d}$	3

A repeated factor needs a fraction for each power of it. Writing

\frac{3 x^{2} + 3 x - 9}{(x + 1)^{2} (x - 2)}

\frac{A}{x + 1} + \frac{B}{x - 2}

cannot work: the identity has no way to produce the right numerator, and the form mark is lost before the algebra starts.

The binomial expansion of (1 + x)^n for a rational n

The expansion works for any rational $n$ , not only for positive integers: $(1 + x)^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + \dots$ , and it is valid only when $| x | < 1$ . When $n$ is a positive integer the series stops after $n + 1$ terms. For every other value of $n$ it continues for ever, so a question asks for the first few terms and expects the coefficients as exact fractions.

To expand $(1 + 3 x)^{1 / 3}$ , use the standard series with $n = \frac{1}{3}$ and then replace $x$ by $3 x$ everywhere. The coefficients are $n = \frac{1}{3}$ , $\frac{n (n - 1)}{2} = \frac{\frac{1}{3} \times (- \frac{2}{3})}{2} = - \frac{1}{9}$ and $\frac{n (n - 1) (n - 2)}{6} = \frac{\frac{1}{3} \times (- \frac{2}{3}) \times (- \frac{5}{3})}{6} = \frac{5}{162}$ . Multiplying those by $(3 x)^{2} = 9 x^{2}$ and $(3 x)^{3} = 27 x^{3}$ gives $1 + x - x^{2} + \frac{5}{6} x^{3} + \dots$ .

The set of values of x for which the expansion is valid comes from the substitution, not from the original variable. The series in $u$ converges when $| u | < 1$ , and here $u = 3 x$ , so the expansion is valid for $| 3 x | < 1$ , that is $| x | < \frac{1}{3}$ , or $- \frac{1}{3} < x < \frac{1}{3}$ . Cambridge asks for this set explicitly, and leaving it out loses a mark.

If the bracket is not of the form $(1 + u)^{n}$ , take the constant out first. $(2 + x)^{- 1} = 2^{- 1} {(1 + \frac{x}{2})}^{- 1} = \frac{1}{2} (1 - \frac{x}{2} + \frac{x^{2}}{4} - \dots) = \frac{1}{2} - \frac{x}{4} + \frac{x^{2}}{8} - \dots$ , valid for $| \frac{x}{2} | < 1$ , that is $| x | < 2$ . Working with exact fractions matters throughout: a decimal such as $0.833 x^{3}$ can lose the accuracy mark where $\frac{5}{6} x^{3}$ scores.

Quote the value of $n$ and the series you are using before you substitute anything
Raise the whole substituted term to the power: $(3 x)^{2} = 9 x^{2}$ , not $3 x^{2}$
Keep the terms in ascending powers of $x$
The validity condition always comes from making the bracket lie between -1 and 1
For $(a + b x)^{n}$ with $a \neq 1$ , factor out $a^{n}$ first so the bracket becomes ${(1 + \frac{b x}{a})}^{n}$

A general term is not examined at 9709, so you are never asked for the term in

x^{10}

. What is examined is the first three or four terms, and the range of values of

x

for which the series is valid.

How a Paper 3 algebra question is put together

Almost every question in this section is a chain rather than a single task. A polynomial is given, you are told something about a factor or a remainder, you find a constant, you factorise, and a later part uses the result. Read the whole question before starting: the constant you are asked for in the first part is usually the one the second part needs.

The wording tells you what the marks are for. Show that means the answer is printed in the question and the marks are for the working that reaches it, so every step must be visible. Determine means the answer has to be found and stated. Sketch means the shape and the intercepts matter and the scale does not. Calculate means the method carries the marks and a bare answer from a calculator will not score.

If a part begins with 'hence', you are expected to use the previous part; a fresh method usually scores nothing
Check whether the numerator is of lower degree than the denominator before you start partial fractions
In a partial-fraction answer, give the constants explicitly as well as the final decomposition
In a binomial question, state the range of values of $x$ in the same answer as the series
Test a partial-fraction answer by putting $x = 0$ into both sides - it takes ten seconds and catches sign errors

Key terms you must be able to define

Modulus: The distance of a number from zero, written $| x |$ , and never negative. $| - 4 | = 4$ .

Modulus function: $y = | a x + b |$ , a V-shaped graph with its vertex on the x-axis where $a x + b = 0$ .

Vertex (of a modulus graph): The corner of the V, at $(- \frac{b}{a}, 0)$ for $y = | a x + b |$ .

Polynomial: A sum of terms of the form $a x^{n}$ with whole-number powers. $2 x^{3} - x + 5$ is a polynomial of degree 3.

Degree: The highest power of $x$ in a polynomial. The degree of $x^{4} - 3 x^{2} + 2 x - 1$ is 4.

Dividend: The polynomial being divided, written on top in a long-division layout.

Divisor: The polynomial you divide by, such as $x - 3$ or $x^{2} + 1$ .

Quotient: The polynomial result of the division, with degree equal to the dividend's degree minus the divisor's degree.

Remainder: What is left over after division. Its degree is less than the degree of the divisor, so a linear divisor leaves a constant.

Division identity: $P (x) = (divisor) \times (quotient) + remainder$ , true for every $x$ .

Remainder theorem: The remainder when $f (x)$ is divided by $x - k$ is $f (k)$ ; dividing by $a x - b$ leaves $f (\frac{b}{a})$ .

Factor theorem: $x - k$ is a factor of $f (x)$ if and only if $f (k) = 0$ .

Root: A value of $x$ for which the polynomial equals zero. A root $k$ means $x - k$ is a factor.

Rational function: A fraction with polynomials above and below the line, such as $\frac{7 x - 1}{(2 x - 1) (x + 2)}$ .

Proper fraction (algebraic): A rational function whose numerator has a lower degree than its denominator. Partial fractions are only asked for in this case.

Partial fractions: The sum of simpler fractions that is identical to a given rational function, such as $\frac{1}{2 x - 1} + \frac{3}{x + 2}$ .

Repeated factor: A factor such as $(x + 1)^{2}$ . It needs a separate fraction for each of $(x + 1)$ and $(x + 1)^{2}$ .

Irreducible quadratic factor: A factor such as $x^{2} + 4$ that cannot be split into real linear factors. Its partial fraction has the form $\frac{B x + C}{x^{2} + 4}$ .

Binomial expansion: $(1 + x)^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + \dots$ for any rational $n$ .

Validity of an expansion: The set of values of $x$ for which the series converges, obtained from $| u | < 1$ in $(1 + u)^{n}$ .

Applied

At AS & A Level the marks sit in the setup: choosing the model, stating the variable and writing the equation before any working. These show the shape of a full-mark answer, then you do the algebra yourself.

Sketching a modulus graph

For $y = | 2 x - 3 |$ , the inside is zero at $x = \frac{3}{2}$ , so the vertex is $(\frac{3}{2}, 0)$ . Putting $x = 0$ gives $y = 3$ , so the graph meets the y-axis at $(0, 3)$ . To the right of the vertex the graph is the line $y = 2 x - 3$ with gradient 2; to the left it is its reflection with gradient $- 2$ . Nothing else is needed: a labelled vertex, a labelled intercept and two straight arms.

Checking a division

Dividing $x^{3} - 4 x^{2} + 2 x + 7$ by $x - 3$ gives quotient $x^{2} - x - 1$ and remainder 4. Check with the identity: $(x - 3) (x^{2} - x - 1) + 4 = x^{3} - x^{2} - x - 3 x^{2} + 3 x + 3 + 4 = x^{3} - 4 x^{2} + 2 x + 7$ . The check also shows why the remainder is the value of the polynomial at $x = 3$ : $27 - 36 + 6 + 7 = 4$ .

Two facts, two unknowns

$f (x) = x^{3} + a x + b$ has $x - 1$ as a factor and leaves a remainder of 10 when divided by $x - 2$ . The factor theorem gives $f (1) = 1 + a + b = 0$ , and the remainder theorem gives $f (2) = 8 + 2 a + b = 10$ . Subtracting: $a = 3$ . Then $b = - 1 - 3 = - 4$ . So $f (x) = x^{3} + 3 x - 4$ , and a quick check gives $f (1) = 0$ and $f (2) = 8 + 6 - 4 = 10$ , as required.

Partial fractions with a quadratic factor

Express $\frac{3 x^{2} + 5 x + 7}{(x + 1) (x^{2} + 4)}$ . The form is $\frac{A}{x + 1} + \frac{B x + C}{x^{2} + 4}$ , so $3 x^{2} + 5 x + 7 = A (x^{2} + 4) + (B x + C) (x + 1)$ . At $x = - 1$ : $5 = 5 A$ , so $A = 1$ . Comparing $x^{2}$ coefficients: $3 = 1 + B$ , so $B = 2$ . Comparing constants: $7 = 4 + C$ , so $C = 3$ . Answer: $\frac{1}{x + 1} + \frac{2 x + 3}{x^{2} + 4}$ . Check at $x = 0$ : left side $\frac{7}{4}$ , right side $1 + \frac{3}{4}$ . Both are $\frac{7}{4}$ .

Stating the validity of an expansion

Expand $(1 + 4 x)^{\frac{1}{2}}$ as far as the term in $x^{3}$ . With $n = \frac{1}{2}$ the coefficients are $\frac{1}{2}$ , $\frac{\frac{1}{2} \times (- \frac{1}{2})}{2} = - \frac{1}{8}$ and $\frac{\frac{1}{2} \times (- \frac{1}{2}) \times (- \frac{3}{2})}{6} = \frac{1}{16}$ . Substituting $4 x$ gives $1 + 2 x - \frac{1}{8} (16 x^{2}) + \frac{1}{16} (64 x^{3}) = 1 + 2 x - 2 x^{2} + 4 x^{3}$ . The series in $u$ needs $| u | < 1$ , and $u = 4 x$ , so the expansion is valid for $| x | < \frac{1}{4}$ .

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What is

| x |

when

x

is negative?

| x | = - x

, which is positive. For

x = - 6

| - 6 | = 6

Q2Where is the vertex of

y = | a x + b |

On the x-axis at

x = - \frac{b}{a}

, the value that makes

a x + b = 0

Q3What is the y-intercept of

y = | 2 x - 3 |

(0, 3)

. Put

x = 0

: the height is

| b |

, which is 3 here.

Q4How do you solve

| a | = | b |

Square both sides:

a^{2} = b^{2}

, then solve the resulting equation and check both answers in the original.

Q5Write

| x - a | < b

without a modulus sign.

a - b < x < a + b

, provided

b > 0

Q6Write

| x - a | > b

without a modulus sign.

x < a - b

x > a + b

, provided

b > 0

Q7When is it unsafe to square an inequality?

Whenever one side might be negative. Square only when both sides are certainly non-negative; otherwise split into cases.

Q8State the division identity for polynomials.

P (x) = (divisor) \times (quotient) + remainder

, with the remainder of lower degree than the divisor.

Q9What is the degree of the remainder when you divide by

x^{2} + 1

At most 1, so the remainder has the form

r x + s

Q10State the remainder theorem.

Dividing

f (x)

x - k

leaves remainder

f (k)

; dividing by

a x - b

leaves

f (\frac{b}{a})

Q11State the factor theorem.

x - k

is a factor of

f (x)

exactly when

f (k) = 0

Q12What form do partial fractions take when the denominator is

(x + 1)^{2} (x - 2)

\frac{A}{x + 1} + \frac{B}{(x + 1)^{2}} + \frac{C}{x - 2}

- one fraction for each power of the repeated factor.

Q13What form do partial fractions take when the denominator is

(x + 1) (x^{2} + 4)

\frac{A}{x + 1} + \frac{B x + C}{x^{2} + 4}

- the quadratic factor takes a linear numerator.

Q14Write the first four terms of

(1 + x)^{n}

for a rational

n

1 + n x + \frac{n (n - 1)}{2} x^{2} + \frac{n (n - 1) (n - 2)}{6} x^{3} + \dots

, valid for

| x | < 1

Q15For which values of

x

is the expansion of

(1 + 3 x)^{\frac{1}{3}}

valid?

| 3 x | < 1

, so

| x | < \frac{1}{3}

. The condition always comes from the substituted term, not from

x

itself.

Q16How do you expand

(2 + x)^{- 1}

Take out the 2 first:

2^{- 1} {(1 + \frac{x}{2})}^{- 1} = \frac{1}{2} - \frac{x}{4} + \frac{x^{2}}{8} - \dots

, valid for

| x | < 2

0 of 16 marked as known

Exam technique for this topic

For a modulus inequality, say which case you are taking before you write any algebra. The mark scheme looks for the split at the value where the inside of the modulus changes sign, and a solution with no split shown loses the method mark even when the answer is right.
When a question gives you a factor or a remainder, substitute immediately and write the resulting equation. 'M1 for using $f (2) = 0$ ' is a mark that only exists on paper - it cannot be awarded from a correct final answer on its own.
In partial fractions, write the form of the answer down as a separate line before you multiply through. It is worth a mark, it prevents the repeated-factor error, and it makes the rest of the working easy to follow.
In a binomial expansion, quote $n$ and the substitution on one line, then the series. Losing the term $(3 x)^{2} = 9 x^{2}$ is the single commonest error in this question, and writing the substitution explicitly prevents it.
Keep every answer exact until the final line. A binomial coefficient of $\frac{5}{6}$ , a logarithm such as $\frac{\ln 5}{\ln 2}$ and a partial fraction of $\frac{2}{3 (2 x + 1)}$ all lose accuracy marks when they are turned into decimals too early.

Why students lose marks here

A student wrote

| x | = x

for every value of

x

, so

| - 4 | = - 4

Why it loses marksThe modulus is a distance, so it can never be negative. The definition has two cases and the second one is the one that removes the sign: for

x < 0

| x | = - x

The answer that scores

| - 4 | = - (- 4) = 4

. More generally

| x | = x

for

x \geq 0

and

| x | = - x

for

x < 0

A student wroteSquaring both sides of

2 x + 5 < | x + 1 |

to get

4 x^{2} + 20 x + 25 < x^{2} + 2 x + 1

Why it loses marksSquaring preserves the direction of an inequality only when both sides are certainly non-negative. The left side here is negative for

x < - 2.5

, so the two inequalities are not equivalent: at

x = - 5

the original reads

- 5 < 4

, which is true, while the squared version reads

25 < 16

, which is false.

The answer that scoresSplit at

x = - 1

. For

x < - 1

2 x + 5 < - x - 1

gives

x < - 2

; for

x \geq - 1

the branch gives

x < - 4

, which is impossible. So

x < - 2

A student wroteThe remainder when

f (x)

is divided by

2 x - 1

f (1)

Why it loses marksThe remainder theorem uses the value that makes the divisor zero. Setting

2 x - 1 = 0

gives

x = \frac{1}{2}

, not

x = 1

The answer that scoresDivide by

2 x - 1

: the remainder is

f (\frac{1}{2})

A student wroteWriting

\frac{3 x^{2} + 3 x - 9}{(x + 1)^{2} (x - 2)}

\frac{A}{x + 1} + \frac{B}{x - 2}

Why it loses marksA repeated factor needs one fraction for each power. Two constants cannot reproduce a numerator of degree 2 against a denominator of degree 3, and the form mark is lost before any algebra is done.

The answer that scoresThe correct form is

\frac{A}{x + 1} + \frac{B}{(x + 1)^{2}} + \frac{C}{x - 2}

, which here gives

\frac{2}{x + 1} + \frac{3}{(x + 1)^{2}} + \frac{1}{x - 2}

A student wroteExpanding

(1 + 3 x)^{\frac{1}{3}}

and stopping at

1 + x

, or leaving the validity as

| x | < 1

Why it loses marksThe first two terms are not 'the expansion' unless the question asks for them, and the validity condition belongs to the substituted variable

u = 3 x

, not to

x

itself.

The answer that scores

1 + x - x^{2} + \frac{5}{6} x^{3} + \dots

, valid for

| 3 x | < 1

, that is

| x | < \frac{1}{3}

Syllabus coverage

The modulus function and the graph of y = |ax + b|

Solving equations and inequalities with modulus

Dividing polynomials: quotient and remainder

The remainder theorem and the factor theorem

Partial fractions: the three denominators Paper 3 uses

The binomial expansion of (1 + x)^n for a rational n

How a Paper 3 algebra question is put together

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here