Algebraic Manipulation

Algebra uses letters to represent unknown or variable quantities. The standard notation conventions are:

Notation	Meaning
$ab$	$a\times b$ (multiplication sign omitted)
$3y$	$y+y+y$, or equivalently $3\times y$
$a^2$	$a\times a$
$a^3$	$a\times a\times a$
$a^{2}b$	$a\times a\times b$
$\frac{a}{b}$	$a÷b$

Coefficients are written as fractions rather than decimals: $\frac{1}{2}x$, not $0.5x$.

Key vocabulary every student must know:

Term	Definition	Example
Term	A single number, variable, or product of both	$3x^2$, $-5$, $2xy$
Factor	A quantity that divides another exactly; can be a number or expression	$3$ and $x$ are factors of $3x$
Expression	A collection of terms — no equals sign	$3x^2+2x-5$
Equation	Two expressions connected by $=$ — can be solved	$3x+1=10$
Formula	An equation expressing one quantity in terms of others	$A=\pi r^2$
Identity	True for all values of the variable; written $\equiv$	$2(x+3)\equiv 2x+6$
Inequality	Expressions connected by $<$, $>$, $\le$, $\ge$	$2x+1>7$

Substitution means replacing letters with given numbers and evaluating the result. Follow BIDMAS strictly.

Worked example — evaluate $5a^2-3b+2$ when $a=-2$ and $b=4$:

$5(-2{)}^2-3(4)+2=5(4)-12+2=20-12+2=10$

Note: $(-2{)}^2=+4$, not $-4$. The bracket ensures the negative sign is squared too.

Worked example — the kinetic energy formula is $E_k=\frac{1}{2}m{v}^2$. Find $E_k$ when $m=3$ and $v=10$:

$E_k=\frac{1}{2}\times 3\times 10^2=\frac{1}{2}\times 3\times 100=150\text{ J}$

Worked example — evaluate $\frac{a+b}{a-b}$ when $a=7$ and $b=3$:

$\frac{7+3}{7-3}=\frac{10}{4}=2.5$

Like terms share exactly the same variable(s) and power(s). Only like terms can be added or subtracted.

Like terms	Unlike terms
$3x$ and $5x$ → $8x$	$3x$ and $5x^2$ (different powers)
$4xy$ and $-xy$ → $3xy$	$4xy$ and $4x$ (different variables)
$7$ and $-3$ → $4$	$7$ and $7x$ (one has a variable)

Worked example — simplify $5x^2+3x-2x^2+7-x$:

Group by type: $x^2$ terms: $5x^2-2x^2=3x^2$; $x$ terms: $3x-x=2x$; constants: $7$

$5x^2+3x-2x^2+7-x=3x^2+2x+7$

Worked example — simplify $4a+3b-2a+5b-ab$:

$=(4a-2a)+(3b+5b)-ab=2a+8b-ab$

Note: $ab$ cannot be combined with $a$ or $b$ terms — it is a separate term.

Single bracket — multiply each term inside by the term outside:

$3(2x-5)=6x-15$

$-2(x^2+3x-1)=-2x^2-6x+2$

Double brackets (FOIL):

$(x+4)(x-3)=x^2-3x+4x-12=x^2+x-12$

$(2x+1)(3x-5)=6x^2-10x+3x-5=6x^2-7x-5$

Perfect squares:

$(x+a{)}^2=x^2+2ax+a^2$

$(x+5{)}^2=x^2+10x+25$ ✓

Expanding three or more brackets (Higher):

$(x+1)(x-2)(x+3)$

Step 1 — expand the first two: $(x+1)(x-2)=x^2-x-2$

Step 2 — expand the result with the third: $(x^2-x-2)(x+3)=x^3+3x^2-x^2-3x-2x-6=x^3+2x^2-5x-6$ ✓

Factorising reverses expansion — it writes an expression as a product.

Taking out a common factor:

$6x^2+9x=3x(2x+3)$

$12a^{2}b-8a{b}^2=4ab(3a-2b)$

Factorising $x^2+bx+c$ — find two numbers that multiply to $c$ and add to $b$:

$x^2+7x+12=(x+3)(x+4)(3\times 4=12;3+4=7)✓$

$x^2-5x+6=(x-2)(x-3)(-2\times -3=6;-2+(-3)=-5)✓$

$x^2+x-12=(x+4)(x-3)(4\times -3=-12;4+(-3)=1)✓$

Difference of two squares:

$a^2-b^2=(a+b)(a-b)$

$x^2-25=(x+5)(x-5)$

$4x^2-9=(2x+3)(2x-3)$

Factorising $a{x}^2+bx+c$ — the AC method: find two numbers that multiply to $ac$ and add to $b$, then split the middle term.

Worked example — factorise $6x^2+7x-3$:

$ac=6\times (-3)=-18$. Find two numbers multiplying to $-18$ and adding to $7$: $9$ and $-2$.

$6x^2+9x-2x-3=3x(2x+3)-1(2x+3)=(3x-1)(2x+3)$

Check: $(3x-1)(2x+3)=6x^2+9x-2x-3=6x^2+7x-3$ ✓

Algebraic fractions (Higher) — simplify by factorising numerator and denominator, then cancel common factors:

$\frac{x^2-x-6}{x^2-9}=\frac{(x-3)(x+2)}{(x-3)(x+3)}=\frac{x+2}{x+3}(x\ne 3)$

Adding algebraic fractions — find a common denominator:

$\frac{2}{x+1}+\frac{3}{x-2}=\frac{2(x-2)+3(x+1)}{(x+1)(x-2)}=\frac{5x-1}{(x+1)(x-2)}$

The index laws apply to algebraic terms exactly as to numbers.

Law	Algebraic form
Multiply	$x^3\times x^4=x^7$
Divide	$x^6÷x^2=x^4$
Power of power	$(x^3{)}^2=x^6$
Zero index	$x^0=1$
Negative index	$x^{-2}=\frac{1}{x^2}$

Simplifying expressions with multiple laws:

$\frac{(2x^3{)}^2\times 3x}{6x^4}=\frac{4x^6\times 3x}{6x^4}=\frac{12x^7}{6x^4}=2x^3$

Surd terms can be collected and manipulated using the same rules as ordinary algebraic terms. A surd like $\sqrt{2}$ acts as an "unknown" — you can add multiples of it.

Collecting like surd terms:

$3\sqrt{2}+5\sqrt{2}=8\sqrt{2}\text{(like terms: both are multiples of }\sqrt{2}\text{)}$

$4\sqrt{3}-\sqrt{3}+2\sqrt{3}=5\sqrt{3}$

$5\sqrt{2}+3\sqrt{3}\text{— cannot be simplified (different surds, unlike terms)}$

Expanding with surd coefficients:

$\sqrt{2}(\sqrt{2}+3)=(\sqrt{2}{)}^2+3\sqrt{2}=2+3\sqrt{2}$

$(1+\sqrt{3})(2-\sqrt{3})=2-\sqrt{3}+2\sqrt{3}-(\sqrt{3}{)}^2=2+\sqrt{3}-3=-1+\sqrt{3}$

Difference of two squares with surds:

$(\sqrt{5}+2)(\sqrt{5}-2)=(\sqrt{5}{)}^2-2^2=5-4=1$

This pattern (rationalising the denominator) is used when surds appear in fractions. See the Surds lesson for simplifying $\sqrt{n}$ itself.

(Extra context — rationalising the denominator by multiplying by the conjugate uses this identity. For example, $\frac{1}{\sqrt{5}+2}=\frac{\sqrt{5}-2}{(\sqrt{5}+2)(\sqrt{5}-2)}=\sqrt{5}-2$.)

1. Not squaring the coefficient when expanding a bracket squared

$(3x{)}^2=9x^2$, not $3x^2$. Both the coefficient and the variable are squared.

2. Sign errors when expanding $(x-a{)}^2$

$(x-5{)}^2=x^2-10x+25$, not $x^2-25$. Squaring a binomial always produces a middle term.

3. Factorising — only taking out a partial common factor

$12x^2+8x=4x(3x+2)$, not $2x(6x+4)$. The factorised form should have no common factor remaining inside the bracket. Take out the highest common factor to ensure no factor remains.

4. Algebraic fractions — cancelling terms, not factors

$\frac{x^2+4}{x+2}$ cannot be simplified to $x+2$ — the numerator does not factorise as $(x+2)(x+2)$. Only cancel common factors (items multiplying the whole numerator or denominator), never individual terms within a sum.

Mistake	Correction
"$(x+3{)}^2=x^2+9$"	$(x+3{)}^2=x^2+6x+9$ (missing middle term $2\times x\times 3$)
"$x^2-16=(x-4{)}^2$"	$x^2-16=(x+4)(x-4)$ (difference of two squares)
"Factorise $x^2+5x+6$: $(x+6)(x-1)$"	$(x+2)(x+3)$: need two numbers multiplying to 6 and adding to 5 → 2 and 3