Algebra: the modulus function and polynomials

The modulus of a number Paper 2

The modulus of a number is its size, ignoring its sign. It is written $\left|x\right|$ and defined by
$$\left|x\right|=x\text{if}x⩾0,\left|x\right|=-x\text{if}x<0$$
So $\left|5\right|=5$ and $|-5|=5$. The second line is not a mistake: if $x$ is negative, $-x$ is positive.

The most useful way to think of $\left|x\right|$ is as a distance: it is the distance of $x$ from 0 on the number line. In the same way, $|x-a|$ is the distance between $x$ and $a$. So $|x-3|=2$ asks for the numbers 2 away from 3, which are 1 and 5, and $|x-3|<2$ asks for the numbers less than 2 away from 3, which is $1<x<5$.

Two facts follow and are used constantly:

• a modulus is never negative, so an equation such as $|2x+1|=-4$ has no solutions;
• $|a{|}^2=a^2$, because squaring removes the sign anyway. Two numbers with the same size have the same square, so $\left|a\right|=\left|b\right|⟺a^2=b^2$.

Graphs of y = |ax + b| Paper 2

To sketch $y=|ax+b|$, first think of the straight line $y=ax+b$. Wherever the line is above the $x$-axis, the modulus changes nothing. Wherever it is below, the modulus makes each $y$ value positive, so that part of the line is reflected in the $x$-axis.

The result is a V shape with its vertex on the $x$-axis, at the root of $ax+b=0$, that is $x=-\frac{b}{a}$. The two arms have gradients $a$ and $-a$. Mark the vertex and the $y$-intercept, $(0,|b|)$, on any sketch.

For $y=|2x-3|$: the vertex is at $(1.5,0)$, the $y$-intercept is $(0,3)$, and the arms have gradients 2 and $-2$.

Only graphs of this linear kind are in Paper 2: graphs of $y=|\mathrm{f}(x)|$ and $y=\mathrm{f}(|x|)$ for non-linear functions are not included.

The part of $y=2x-3$ below the $x$-axis is reflected upwards to make the V of $y=|2x-3|$.

Solving modulus equations Paper 2

There are two reliable methods for an equation with a modulus on each side, $|\mathrm{f}(x)|=|\mathrm{g}(x)|$ with $\mathrm{f}$ and $\mathrm{g}$ linear.

Squaring. Since $\left|a\right|=\left|b\right|⟺a^2=b^2$, square both sides and solve the quadratic. Squaring is safe here because both sides are already non-negative, so no false solutions appear.

Two cases. Two numbers have the same modulus when they are equal or when one is the negative of the other. So solve $\mathrm{f}(x)=\mathrm{g}(x)$ and $\mathrm{f}(x)=-\mathrm{g}(x)$. This is often quicker and avoids a quadratic.

When one side has no modulus, as in $|2x-1|=3x+4$, the two cases still work, but check every answer in the original equation: the right-hand side must not be negative, because a modulus cannot equal a negative number. Squaring such an equation can introduce a false root, so the cases method with a check is the safer choice.

Solving modulus inequalities Paper 2

The distance idea gives the basic results at once. For $b>0$:

• $|x-a|<b⟺a-b<x<a+b$ (within $b$ of $a$);
• $|x-a|>b⟺x<a-b$ or $x>a+b$ (more than $b$ away from $a$).

For $|2x-7|⩽5$, write $-5⩽2x-7⩽5$, add 7 to get $2⩽2x⩽12$, and halve: $1⩽x⩽6$.

Modulus on both sides. For $|\mathrm{f}(x)|<|\mathrm{g}(x)|$, both sides are non-negative, so squaring keeps the inequality true: solve $\mathrm{f}(x{)}^2<\mathrm{g}(x{)}^2$ as a quadratic inequality, using the critical values and a sketch of the parabola.

Modulus on one side only. For an inequality such as $3x-1<|x+3|$, do not square: the left side can be negative, and squaring a negative number reverses its size order. Instead, find where the graphs meet and use a sketch to see where one graph is above the other. Where the line $y=3x-1$ is below the $x$-axis it is certainly below the modulus graph, which a squared inequality would miss.

The line lies below the V for every $x<2$, including where the line is below the $x$-axis.

Polynomials and long division Paper 2

A polynomial in $x$ is a sum of terms $a{x}^n$ with $n$ a non-negative whole number, such as $2x^3+5x^2-4x+1$. Its degree is the highest power: 3 here, so it is a cubic.

Dividing a polynomial $\mathrm{p}(x)$ by a divisor $\mathrm{d}(x)$ means writing
$$\mathrm{p}(x)\equiv \mathrm{d}(x)\mathrm{q}(x)+\mathrm{r}(x)$$
where $\mathrm{q}(x)$ is the quotient and the remainder $\mathrm{r}(x)$ has a lower degree than $\mathrm{d}(x)$. Dividing by a linear divisor leaves a constant remainder; dividing by a quadratic leaves a remainder of the form $Ax+B$, which may be zero. This is exactly the same idea as $47=6\times 7+5$ in arithmetic.

Long division builds the quotient one term at a time:

1. write both polynomials in descending powers, with a $0$ term for any missing power;
2. divide the leading term of what is left by the leading term of the divisor: that is the next term of the quotient;
3. multiply the whole divisor by that term and subtract;
4. repeat until what is left has a lower degree than the divisor.

Each step removes the highest power of what is left, which is why the process must end. Always check by multiplying back: $\mathrm{d}(x)\mathrm{q}(x)+\mathrm{r}(x)$ must give $\mathrm{p}(x)$.

The remainder theorem Paper 2

Write the division by $(x-a)$ as $\mathrm{p}(x)\equiv (x-a)\mathrm{q}(x)+R$, where $R$ is a constant. Substitute $x=a$: the first term becomes $0\times \mathrm{q}(a)=0$, so $\mathrm{p}(a)=R$.

The remainder theorem: when $\mathrm{p}(x)$ is divided by $(x-a)$, the remainder is $\mathrm{p}(a)$.

For a divisor $(ax+b)$, the same argument with $x=-\frac{b}{a}$, the value that makes the divisor zero, gives the remainder $\mathrm{p}(-\frac{b}{a})$. So the remainder when $4x^3-6x^2+5$ is divided by $(2x-1)$ is $\mathrm{p}(\frac{1}{2})=\frac{1}{2}-\frac{3}{2}+5=4$, agreeing with the long division above.

The theorem gives the remainder without dividing, which is why it is the tool for finding an unknown coefficient: a given remainder becomes an equation in the unknown. Be careful with the sign: dividing by $(x+2)$ means substituting $x=-2$.

The factor theorem Paper 2

A remainder of zero means the divisor goes exactly. So:

The factor theorem: $(x-a)$ is a factor of $\mathrm{p}(x)$ if and only if $\mathrm{p}(a)=0$; and $(ax+b)$ is a factor if and only if $\mathrm{p}(-\frac{b}{a})=0$.

To factorise a cubic, search for a root among the factors of the constant term (and fractions such as $\frac{1}{2}$ or $\frac{1}{3}$ when the leading coefficient is not 1). Once $\mathrm{p}(a)=0$, divide by $(x-a)$, or compare coefficients, to find the quadratic factor, then factorise that.

To show that $(x-a)$ is a factor, the examiner wants to see the substitution worked out and the conclusion: "$\mathrm{p}(3)=54-27-33+6=0$, so $(x-3)$ is a factor."

Unknown coefficients. Each piece of information (a factor, or a remainder) gives one equation. Two unknowns need two pieces of information, which give simultaneous equations.