Algebraic Manipulation – OCR FSMQ Additional Maths

1 Vocabulary and Notation

AL1 asks you to know and use algebraic vocabulary and notation. The specification gives a complete list, which means each of these words can be asked about directly: constant, coefficient, expression, equation, identity, index, variable, unknown, and $\mathrm{f}(x)$.

Term	What it means	In $\mathrm{f}(x) = 4x^3 - x + 9$
Constant	A number that does not change	$9$
Coefficient	The number multiplying a power of the variable	$4$ is the coefficient of $x^3$; $-1$ of $x$
Index	A power	$3$ and $1$
Variable	The quantity free to change	$x$
Expression	Terms with no equals sign	$4x^3 - x + 9$
Function notation	$\mathrm{f}(x)$ is the output for input $x$	$\mathrm{f}(2) = 32 - 2 + 9 = 39$

Equation vs identity

An equation holds for particular values: $2x = 6$ only when $x = 3$
An identity holds for every value: $2(x+1) \equiv 2x + 2$

Worked Example 1 — Using function notation

$\mathrm{f}(x) = x^2 - 3x$. Find $\mathrm{f}(4)$, $\mathrm{f}(-1)$ and $\mathrm{f}(a+1)$.

①$\mathrm{f}(4) = 16 - 12 = 4$

②$\mathrm{f}(-1) = 1 - (-3) = 1 + 3 = 4$

③$\mathrm{f}(a+1) = (a+1)^2 - 3(a+1)$

④$= a^2 + 2a + 1 - 3a - 3 = a^2 - a - 2$

Substituting a bracket into a function is exactly how the factor theorem is set up later. Get comfortable with it now.

Watch the negative. $\mathrm{f}(-1)$ with $\mathrm{f}(x) = x^2 - 3x$ gives $(-1)^2 = +1$, and $-3 \times (-1) = +3$. Both signs catch people out.

2 Algebraic Fractions

AL2 requires you to simplify expressions involving algebraic fractions. The specification's own example is $\dfrac{1}{x-1} - \dfrac{2}{x+1}$.

Factorise every numerator and denominator that will factorise.
For adding or subtracting, find the lowest common denominator.
Combine over the single denominator, expanding carefully.
Cancel any factor common to the whole numerator and denominator.

Worked Example 2 — Subtracting two fractions

Simplify $\dfrac{1}{x-1} - \dfrac{2}{x+1}$.

①Common denominator is $(x-1)(x+1)$.

②$\dfrac{(x+1) - 2(x-1)}{(x-1)(x+1)}$

③Numerator: $x + 1 - 2x + 2 = 3 - x$

④$= \dfrac{3-x}{(x-1)(x+1)}$ or equivalently $\dfrac{3-x}{x^2-1}$

The bracket matters. Subtracting $2(x-1)$ changes both signs: $-2x + 2$, not $-2x - 2$.

Worked Example 3 — Cancelling after factorising

Simplify $\dfrac{x^2 - 3x}{x^2 - x - 6}$.

①Numerator: $x^2 - 3x = x(x-3)$

②Denominator: $x^2 - x - 6 = (x-3)(x+2)$

③$\dfrac{x(x-3)}{(x-3)(x+2)}$ — the factor $(x-3)$ cancels.

④$= \dfrac{x}{x+2}$

Worked Example 4 — Dividing fractions

Simplify $\dfrac{x^2 - 4}{3x} \div \dfrac{x+2}{6x^2}$.

①Dividing means multiplying by the reciprocal:

②$\dfrac{x^2-4}{3x} \times \dfrac{6x^2}{x+2}$

③$x^2 - 4 = (x-2)(x+2)$, so $(x+2)$ cancels.

④$\dfrac{(x-2) \times 6x^2}{3x} = 2x(x-2)$

You may never cancel across a $+$ or $-$. $\dfrac{x+4}{4} \neq x$, and $\dfrac{x^2+1}{x} \neq x + 1$. Factorise first — if it will not factorise, it will not cancel.

3 Indices

The index laws

$x^a \times x^b = x^{a+b}$ · $\dfrac{x^a}{x^b} = x^{a-b}$ · $(x^a)^b = x^{ab}$
$x^0 = 1$ · $x^{-a} = \dfrac{1}{x^a}$ · $x^{1/n} = \sqrt[n]{x}$ · $x^{m/n} = \left(\sqrt[n]{x}\right)^m$

Worked Example 5 — Mixed indices

Simplify (a) $\dfrac{6x^5 \times 2x^{-3}}{4x}$, (b) $\left(8x^6\right)^{2/3}$.

①(a) Numbers: $\dfrac{6 \times 2}{4} = 3$

②Powers: $x^{5 + (-3) - 1} = x^{1}$, so the answer is $3x$.

③(b) $\left(8x^6\right)^{2/3} = 8^{2/3} \times x^{6 \times 2/3}$

④$8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$, and $x^4$.

⑤$= 4x^4$

For a fractional power, take the root first (denominator) then the power (numerator). $8^{2/3}$ is much easier as $2^2$ than as $\sqrt[3]{64}$.

4 Square Roots and Surds

AL2 also covers square roots. The specification's examples are $\sqrt{125}$, $\sqrt{12} + \sqrt{27}$ and $\dfrac{1}{2 + \sqrt{3}}$ — one of each of the three standard tasks.

The surd rules

$\sqrt{ab} = \sqrt{a}\,\sqrt{b}$ · $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$ · $\left(\sqrt{a}\right)^2 = a$
There is no rule for $\sqrt{a+b}$ — it is not $\sqrt a + \sqrt b$.

Worked Example 6 — Simplifying and collecting

Simplify (a) $\sqrt{125}$, (b) $\sqrt{12} + \sqrt{27}$.

①(a) Look for the largest square factor: $125 = 25 \times 5$.

②$\sqrt{125} = \sqrt{25}\sqrt{5} = 5\sqrt{5}$

③(b) $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

④$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$

⑤Now they are like terms: $2\sqrt3 + 3\sqrt3 = 5\sqrt{3}$

You can only add surds once they share the same surd part. Always simplify each one first.

Rationalising the denominator

Multiply top and bottom by the conjugate:
for $a + \sqrt{b}$ use $a - \sqrt{b}$, because $(a+\sqrt b)(a-\sqrt b) = a^2 - b$

Worked Example 7 — Rationalising

Express $\dfrac{1}{2 + \sqrt{3}}$ with a rational denominator.

①The conjugate of $2 + \sqrt3$ is $2 - \sqrt3$.

②$\dfrac{1}{2+\sqrt3} \times \dfrac{2-\sqrt3}{2-\sqrt3} = \dfrac{2-\sqrt3}{(2+\sqrt3)(2-\sqrt3)}$

③Denominator: $2^2 - \left(\sqrt3\right)^2 = 4 - 3 = 1$

④$= 2 - \sqrt{3}$

The denominator becoming $1$ is a lucky accident here, but the method is always the same. Check your conjugate by confirming the surd cancels.

5 Quick Reference

Coefficient

Includes its sign: in $5-7x$ it is $-7$.

Identity

Use $\equiv$ when true for every $x$.

$\mathrm{f}(a+1)$

Substitute the whole bracket, then expand.

Adding fractions

LCD first; bracket the numerators.

Cancelling

Only whole factors — never across $+$ or $-$.

Dividing fractions

Multiply by the reciprocal.

$x^{m/n}$

Root first, then power.

Simplifying surds

Pull out the largest square factor.

Rationalising

Multiply by the conjugate $a - \sqrt b$.

No such rule

$\sqrt{a+b} \neq \sqrt a + \sqrt b$.

6 Practice Questions

Question 1

$\mathrm{f}(x) = 2x^2 - 5x + 1$. Find $\mathrm{f}(3)$ and $\mathrm{f}(-2)$.

▶ Show solution

$\mathrm{f}(3) = 2(9) - 15 + 1 = 18 - 15 + 1 = 4$

$\mathrm{f}(-2) = 2(4) - 5(-2) + 1 = 8 + 10 + 1 = 19$

Question 2

State whether each of these is an equation or an identity: (a) $3(x - 2) = 3x - 6$, (b) $3x - 2 = 7$.

▶ Show solution

(a) An identity — expanding the left side gives the right side for every $x$. It should be written $3(x-2) \equiv 3x - 6$.

(b) An equation — only true when $x = 3$.

Question 3

Simplify $\dfrac{3}{x} + \dfrac{2}{x-2}$.

▶ Show solution

Common denominator $x(x-2)$:

$\dfrac{3(x-2) + 2x}{x(x-2)} = \dfrac{3x - 6 + 2x}{x(x-2)} = \dfrac{5x-6}{x(x-2)}$

Question 4

Simplify $\dfrac{x^2 + 5x + 6}{x^2 - 9}$.

▶ Show solution

Numerator: $(x+2)(x+3)$

Denominator: $(x-3)(x+3)$

$(x+3)$ cancels: $\dfrac{x+2}{x-3}$

Question 5

Simplify $\dfrac{10x^4 \times 3x^{-2}}{6x^3}$.

▶ Show solution

Numbers: $\dfrac{10 \times 3}{6} = 5$

Powers: $x^{4 + (-2) - 3} = x^{-1}$

$= 5x^{-1} = \dfrac{5}{x}$

Question 6

Evaluate (a) $16^{3/4}$, (b) $27^{-2/3}$.

▶ Show solution

(a) $\sqrt[4]{16} = 2$, then $2^3 = 8$.

(b) $\sqrt[3]{27} = 3$, then $3^2 = 9$, and the negative index gives the reciprocal: $\dfrac{1}{9}$.

Question 7

Express $\sqrt{50} - \sqrt{18}$ in the form $k\sqrt{2}$.

▶ Show solution

$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt2$

$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt2$

$5\sqrt2 - 3\sqrt2 = 2\sqrt{2}$, so $k = 2$.

Question 8

Express $\dfrac{6}{\sqrt{5} - 1}$ with a rational denominator.

▶ Show solution

Multiply by the conjugate $\sqrt5 + 1$:

$\dfrac{6(\sqrt5+1)}{(\sqrt5-1)(\sqrt5+1)} = \dfrac{6(\sqrt5+1)}{5 - 1} = \dfrac{6(\sqrt5+1)}{4}$

$= \dfrac{3(\sqrt5+1)}{2}$ or $\dfrac{3\sqrt5 + 3}{2}$

Question 9

$\mathrm{f}(x) = x^2 + x$. Show that $\mathrm{f}(x+1) - \mathrm{f}(x) = 2x + 2$.

▶ Show solution

$\mathrm{f}(x+1) = (x+1)^2 + (x+1) = x^2 + 2x + 1 + x + 1 = x^2 + 3x + 2$

$\mathrm{f}(x+1) - \mathrm{f}(x) = \left(x^2 + 3x + 2\right) - \left(x^2 + x\right)$

$= 2x + 2$ as required.

Question 10

A right-angled triangle has legs of length $1 + \sqrt{3}$ and $1 - \sqrt{3}$ … or so a student claims.

(a) Explain why those two lengths cannot both be sides of a triangle. (b) A rectangle has width $2 + \sqrt{5}$ and length $2 - \sqrt{5} + 2\sqrt{5}$. Find its exact area, simplified. (c) Show that $\dfrac{1}{1+\sqrt2} + \dfrac{1}{1-\sqrt2} = -2$.

▶ Show solution

(a) $\sqrt3 \approx 1.732$, so $1 - \sqrt3 \approx -0.732$, which is negative. A length cannot be negative, so this is impossible.

(b) First simplify the length: $2 - \sqrt5 + 2\sqrt5 = 2 + \sqrt5$.

Area $= (2+\sqrt5)(2+\sqrt5) = (2+\sqrt5)^2$

$= 4 + 4\sqrt5 + 5 = 9 + 4\sqrt{5}$

(c) Common denominator $(1+\sqrt2)(1-\sqrt2) = 1 - 2 = -1$.

Numerator: $(1-\sqrt2) + (1+\sqrt2) = 2$

So the sum is $\dfrac{2}{-1} = -2$ ✓

Note how the surds cancelled in the numerator while the denominator became rational on its own — that is the conjugate pair at work.

Algebraic Manipulation (AL1–AL2) · OCR FSMQ Additional Maths · Created with MathJax