Algebra

The modulus function Paper 3

The modulus of $x$, written $\left|x\right|$, is its size without its sign: $\left|5\right|=5$ and $|-5|=5$. Formally, $\left|x\right|=x$ when $x⩾0$ and $\left|x\right|=-x$ when $x<0$. Geometrically, $|x-a|$ is the distance between $x$ and $a$ on the number line, and that one idea explains every rule below.

The graph of $y=|ax+b|$. Draw the straight line $y=ax+b$, then reflect in the $x$-axis every part that lies below it. The result is a V shape with its vertex on the $x$-axis at $x=-\frac{b}{a}$, where $ax+b=0$. The two arms have gradients $a$ and $-a$. Always mark the vertex and the $y$-intercept, $(0,|b|)$.

On this paper only the graphs of $y=|ax+b|$ with a linear inside are needed: graphs of $y=|\mathrm{f}(x)|$ and $y=\mathrm{f}(|x|)$ for non-linear $\mathrm{f}$ are not included.

The part of $y=2x-1$ below the axis (dashed) is reflected to make the left arm.

Solving modulus equations and inequalities Paper 3

Three standard tools, each following from the meaning of modulus.

• Two moduli equal. $\left|a\right|=\left|b\right|⟺a^2=b^2$, because both sides are non-negative. Equivalently $a=b$ or $a=-b$. Squaring turns $|3x-2|=|2x+7|$ into a quadratic and finds both solutions at once.
• Modulus less than a number. $|x-a|<b⟺a-b<x<a+b$: $x$ is within distance $b$ of $a$. Similarly $|x-a|>b$ means $x<a-b$ or $x>a+b$, two separate intervals.
• Modulus against a line. For something like $2x+5<|x+1|$, sketch both graphs, find where they meet by solving the two linear cases, and read off where one graph is below the other. The sketch matters: one of the two cases may give a point that is not on the graph at all.

Checking. When you square, or split into cases, check every answer in the original equation. Squaring both sides is safe only when both sides are known to be non-negative, which is why it is the right tool for $\left|a\right|=\left|b\right|$ and the wrong one for $|2x-3|=4-x$ until you know $4-x⩾0$.

Inequalities by squaring. $\left|a\right|<\left|b\right|⟺a^2<b^2$, so a modulus inequality on both sides becomes a quadratic inequality. Find the critical values, then decide which side of them is wanted.

Dividing polynomials Paper 3

Dividing a polynomial $\mathrm{p}(x)$ by a divisor $\mathrm{d}(x)$ gives a quotient $\mathrm{q}(x)$ and a remainder $\mathrm{r}(x)$ with
$$\mathrm{p}(x)\equiv \mathrm{d}(x)\mathrm{q}(x)+\mathrm{r}(x)$$
where the remainder has lower degree than the divisor. Dividing by a linear divisor leaves a constant remainder; dividing by a quadratic leaves a remainder of the form $ax+b$. On this paper $\mathrm{p}(x)$ has degree at most 4.

Long division works exactly like number long division. At each step, divide the leading term of what is left by the leading term of the divisor, write that term in the quotient, multiply the whole divisor by it, and subtract. Stop when what is left has lower degree than the divisor.

Two habits prevent most errors: write in a $0x^k$ for every missing power before you start, and subtract the whole line by changing every sign. The identity above is also the quickest check: multiply out $\mathrm{d}(x)\mathrm{q}(x)+\mathrm{r}(x)$ and you should get $\mathrm{p}(x)$ back.

The factor and remainder theorems Paper 3

Put $x=a$ into $\mathrm{p}(x)\equiv (x-a)\mathrm{q}(x)+R$. The first term vanishes, leaving $\mathrm{p}(a)=R$. This is the remainder theorem: the remainder when $\mathrm{p}(x)$ is divided by $(x-a)$ is $\mathrm{p}(a)$.

For a divisor $(ax+b)$ with $a\ne 1$, substitute the value that makes it zero: the remainder is $\mathrm{p}(-\frac{b}{a})$. Dividing by $(2x-1)$ means evaluating $\mathrm{p}(\frac{1}{2})$, not $\mathrm{p}(1)$ or $\mathrm{p}(-1)$.

The factor theorem is the special case $R=0$: $(ax+b)$ is a factor of $\mathrm{p}(x)$ exactly when $\mathrm{p}(-\frac{b}{a})=0$.

These theorems are used to:

• find factors and so solve polynomial equations: try small values such as $\pm 1,\pm 2,\pm \frac{1}{2}$, then divide out the factor found and factorise the quadratic quotient;
• find unknown coefficients: each piece of information (a factor, or a remainder) gives one equation, so two unknowns need two facts and simultaneous equations;
• calculate remainders without doing the division.

Partial fractions Paper 3

Adding fractions combines them over a common denominator; partial fractions reverse that, splitting a fraction into simpler ones. It is the step that makes later work (expanding as a series, and integrating) possible.

The form depends only on the denominator. You must recall it, because questions often do not give it:

• Distinct linear factors: $\frac{\dots }{(ax+b)(cx+d)(ex+f)}\equiv \frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{ex+f}$
• A repeated factor: $\frac{\dots }{(ax+b)(cx+d{)}^2}\equiv \frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{(cx+d{)}^2}$
• A quadratic factor $c{x}^2+d$ that does not factorise: $\frac{\dots }{(ax+b)(c{x}^2+d)}\equiv \frac{A}{ax+b}+\frac{Bx+C}{c{x}^2+d}$

Why these forms? The number of unknown constants always equals the degree of the denominator, which is what is needed to match every coefficient of the numerator. A quadratic denominator needs a linear numerator $Bx+C$, and a repeated factor needs both $\frac{B}{cx+d}$ and $\frac{C}{(cx+d{)}^2}$, because $\frac{C}{(cx+d{)}^2}$ alone cannot produce every numerator.

The method. Multiply both sides by the denominator to get an identity, true for every $x$. Then

• substitute the values of $x$ that make each linear factor zero, which wipes out all but one constant at a time;
• compare coefficients (often of $x^2$, or the constant term) for any constant left.

Degree. The method above needs the numerator's degree to be less than the denominator's. Cases where the numerator's degree is greater are excluded on this paper. If the degrees are equal, there is also a constant term: $\frac{3x^2+2x-2}{(x-1)(x+2)}\equiv A+\frac{B}{x-1}+\frac{C}{x+2}$, with $A$ found by comparing the $x^2$ coefficients.

The binomial expansion for any rational power Paper 3

For a positive whole number $n$ the expansion of $(1+x{)}^n$ stops after $n+1$ terms. For any rational $n$ (negative, or a fraction) the same pattern never stops:
$$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+\dots$$
This is given in the list of formulae (MF19). It is an infinite series, so it equals $(1+x{)}^n$ only when the terms shrink fast enough, which is when $\left|x\right|<1$. Outside that interval the series does not settle to any value. Questions ask for the first few terms, so finding the general term is not needed.

Substituting. Replace $x$ by the whole term, in brackets. For $(1-2x{)}^{-3}$, use $n=-3$ and "$x$" $=-2x$: the $x^2$ term is $\frac{(-3)(-4)}{2}(-2x{)}^2=6\times 4x^2=24x^2$. Most errors are a lost minus sign or forgetting to square the 2.

Adapting the series. For $(a+bx{)}^n$, take out $a^n$ first:
$$(a+bx{)}^n=a^n(1+\frac{b}{a}x{)}^n$$
then expand. This is valid when $\left|\frac{b}{a}x\right|<1$, that is $\left|x\right|<\left|\frac{a}{b}\right|$. For example $(2-\frac{1}{2}x{)}^{-1}=\frac{1}{2}(1-\frac{1}{4}x{)}^{-1}=\frac{1}{2}(1+\frac{1}{4}x+\frac{1}{16}{x}^2+\dots )$, valid for $\left|x\right|<4$.

Combining. A product such as $(1-2x)(1+3x{)}^{-2}$ is expanded by expanding the bracket with the rational power, then multiplying, keeping only the powers asked for. A sum of partial fractions is expanded term by term; the whole expansion is valid where every term's expansion is valid, so take the narrowest interval.