Algebraic Manipulation Notes

Q: How do you factorise quadratics?

Find two numbers that multiply to c and add to b. For x^2 + bx + c find two numbers that multiply to c and add to b. So x^2 + 5x + 6 = (x + 2)(x + 3) and x^2 - 3x - 10 = (x - 5)(x + 2).

AQA GCSE Maths: Revision notes

Key facts

$ab$ means $a \times b$ , $2a$ means $2 \times a$ , $a^2$ means $a \times a$ .
Only like terms (same letters and powers) can be collected.
Expand by multiplying every term; factorise by taking out the highest common factor.
Factorise quadratics $x^2 + bx + c$ with two numbers that multiply to $c$ and add to $b$ .
Rearrange formulae with inverse operations; collect and factorise if the subject appears twice (Higher).

Grid for (x + 3)(x + 2), drawn with x = 4: x² + 3x + 2x + 6.

Algebraic notation

Algebraic shorthand leaves out multiplication signs, and follows the order of operations when substituting.

A term is a part separated by $+$ or $-$ : $3x + 2y - 5$ has three terms. The number in front of a letter is the coefficient.

Like terms have the same letters and powers, so $3x$ and $5x$ are like terms but $3x$ and $3x^2$ are not.

When substituting, follow BIDMAS: brackets, powers, then multiply and divide, then add and subtract.

1
Brackets
2
Indices
powers
3
Divide and multiply
left to right
4
Add and subtract
left to right

The order of operations (BIDMAS) when substituting.

$ab$ $a \times b$
$2a$ $2 \times a$
$a^2$ $a \times a$

Worked example

Evaluate $3x^2 - 2x$ when $x = -2$ .

1
Substitute: $3 \times (-2)^2 - 2 \times (-2)$ .
2
Powers: $(-2)^2 = 4$ , so $3 \times 4 = 12$ .
3
$-2 \times (-2) = +4$ , so $12 + 4$ .

What does $3x^2$ mean?

Collecting like terms

Add or subtract the coefficients of like terms and leave the unlike terms alone.

Collecting like terms means adding and subtracting terms with identical letter parts.

Identify the like terms.
Add or subtract their coefficients.
Write the simplified answer.

$3x + 5y + 2x - y = (3x + 2x) + (5y - y) = 5x + 4y$

1
Spot the like terms
3x and 2x; 5y and −y
2
Add the coefficients
3 + 2 = 5 and 5 − 1 = 4
3
Write the answer
5x + 4y

Collecting like terms in 3x + 5y + 2x − y.

Like terms (can combine)

$3x$ and $7x$ give $10x$
$2xy$ and $-5xy$ give $-3xy$
$4a^2$ and $a^2$ give $5a^2$

Unlike terms (cannot combine)

$3x$ and $3x^2$ (different powers)
$2a$ and $3b$ (different letters)
$5$ and $5x$ (no letter on one)

Worked example

Simplify $4a + 3b - 2a + b + 5$ .

1
$a$ terms: $4a - 2a = 2a$ .
2
$b$ terms: $3b + b = 4b$ .
3
The constant $5$ stays.

Simplify $5x + 2y - 3x$ .

Expanding and factorising

Expanding multiplies every term out; factorising writes the highest common factor outside brackets.

Expand a single bracket by multiplying every term inside: $3(2x + 5) = 6x + 15$ and $-2(a - 3) = -2a + 6$ .

For two brackets, multiply each term by each term: $(x + 3)(x + 2) = x^2 + 5x + 6$ and $(2x - 1)(x + 4) = 2x^2 + 7x - 4$ .

Factorise by taking out the HCF: $6x + 9 = 3(2x + 3)$ and $12a^2b + 8ab = 4ab(3a + 2)$ .

Grid for (x + 3)(x + 2), drawn with x = 4: x² + 3x + 2x + 6.

Worked example

Factorise $15xy + 10x^2$ .

1
The HCF is $5x$ .
2
Divide each term: $15xy \div 5x = 3y$ and $10x^2 \div 5x = 2x$ .
3
Check: $5x \times 3y + 5x \times 2x$ .

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

Factorise $x^2 - 9$ (Higher).

Factorising quadratics

Find two numbers that multiply to $c$ and add to $b$ .

For $x^2 + bx + c$ find two numbers that multiply to $c$ and add to $b$ . So $x^2 + 5x + 6 = (x + 2)(x + 3)$ and $x^2 - 3x - 10 = (x - 5)(x + 2)$ .

For $ax^2 + bx + c$ (Higher), find two numbers that multiply to $ac$ and add to $b$ , split the middle term, then factorise in pairs.

Completing the square writes $x^2 + bx + c$ as $(x + \frac{b}{2})^2 - (\frac{b}{2})^2 + c$ .

y = (x + 1)(x + 5) = (x + 3)² − 4: roots and turning point.

Worked example

Complete the square for $x^2 + 8x - 3$ .

1
Half of 8 is 4, so $x^2 + 8x = (x + 4)^2 - 16$ .
2
Bring back the $-3$ .

Factorise $x^2 + 7x + 12$ .

Algebraic fractions

Factorise top and bottom, then cancel common factors (Higher).

Factorise the numerator and denominator, then cancel common factors: $\frac{3x + 6}{x + 2} = \frac{3(x + 2)}{x + 2} = 3$ for $x \neq -2$ .

To add fractions use a common denominator: $\frac{2}{x} + \frac{3}{y} = \frac{2y + 3x}{xy}$ .

State any value that would divide by zero.

(x² − 4) ÷ (x − 2) simplifies to x + 2, but the original is undefined at x = 2.

$\dfrac{x^2 - 4}{x - 2}$ $x + 2$
$\dfrac{a}{b} + \dfrac{c}{d}$ $\dfrac{ad + bc}{bd}$

Simplify $\dfrac{x^2 - 4}{x - 2}$ .

Rearranging formulae

Undo each operation with its inverse, collecting and factorising if the subject appears twice.

The subject is the letter on its own. If it is added, subtract; if multiplied, divide.

$v = u + at$ gives $a = \frac{v - u}{t}$
$v^2 = u^2 + 2as$ gives $a = \frac{v^2 - u^2}{2s}$

If the subject is on both sides, collect those terms and factorise: $ax + b = cx + d$ gives $x(a - c) = d - b$ , so $x = \frac{d - b}{a - c}$ .

1
Collect
the terms containing the subject on one side
2
Factorise
if the subject appears more than once
3
Undo
each operation using its inverse
4
Check
the subject is alone

Making a letter the subject

Worked example

Rearrange $T = 2\pi\sqrt{\dfrac{L}{g}}$ to make $g$ the subject.

1
Square both sides: $T^2 = 4\pi^2 \times \dfrac{L}{g}$ .
2
Multiply by $g$ : $gT^2 = 4\pi^2 L$ .
3
Divide by $T^2$ .

Make $x$ the subject of $y = mx + c$ .

Try an exam question

Factorise fully $12a^2b + 8ab$ , and make $a$ the subject of $v^2 = u^2 + 2as$ .

[4 marks]

That's the notes covered.

Carry on to the next subtopic.