Algebraic techniques - Year 10 Mathematics

What you will learn

factorise expressions by taking out the highest common algebraic factor,
simplify products and quotients using exponent laws,
add, subtract, multiply and divide algebraic fractions,
expand binomial products and perfect squares,
factorise quadratics using grouping, difference of two squares and completing the square,
substitute into formulas and rearrange to make a different variable the subject.

Worked example 0 Real-world example: rearranging a physics formula

The kinetic energy formula is $E = \dfrac{1}{2}mv^2$ . A vehicle of mass $1200$ kg has kinetic energy $48\,000$ J. Find its speed.

Rearrange for $v$ : multiply both sides by $2$ : $2E = mv^2$ .
Divide by $m$ : $v^2 = \dfrac{2E}{m}$ .
Take the square root: $v = \sqrt{\dfrac{2E}{m}}$ .
Substitute: $v = \sqrt{\dfrac{2 \times 48\,000}{1200}} = \sqrt{80} = 4\sqrt{5} \approx 8.94$ m/s.

Key idea: rearranging before substituting keeps the algebra clean and avoids rounding errors.

1. Factorising by common algebraic factors

To factorise, find the highest common factor (HCF) of all terms — including both numerical and algebraic parts — and write it out the front.

Formula reference

ab + ac = a(b + c)

Worked example 1 Taking out a common factor

Factorise $6x^3y - 9x^2y^2 + 3xy$ .

Numerical HCF: $\gcd(6, 9, 3) = 3$ .
Variable HCF: $x$ (lowest power) and $y$ (lowest power) $= xy$ .
HCF $= 3xy$ . Divide each term: $\dfrac{6x^3y}{3xy} = 2x^2$ , $\dfrac{-9x^2y^2}{3xy} = -3xy$ , $\dfrac{3xy}{3xy} = 1$ .
Result: $3xy(2x^2 - 3xy + 1)$ .

2. Exponent laws for products and quotients

Formula reference

a^m \times a^n = a^{m+n} \qquad \frac{a^m}{a^n} = a^{m-n} \qquad (a^m)^n = a^{mn}

(ab)^n = a^n b^n \qquad \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \qquad a^0 = 1 \;(a \neq 0)

Worked example 2 Simplifying with exponent laws

Simplify $\dfrac{12a^5 b^3}{4a^2 b^7}$ .

Coefficients: $\dfrac{12}{4} = 3$ .
$a$ terms: $a^{5-2} = a^3$ .
$b$ terms: $b^{3-7} = b^{-4} = \dfrac{1}{b^4}$ .
Result: $\dfrac{3a^3}{b^4}$ .

Worked example 3 Simplifying a product of powers

Simplify $(2x^3)^4 \times 3x^{-2}$ .

$(2x^3)^4 = 2^4 \cdot x^{12} = 16x^{12}$ .
$16x^{12} \times 3x^{-2} = 48x^{10}$ .

3. Algebraic fractions

Algebraic fractions follow the same rules as numerical fractions. Always factorise first to identify common factors that cancel.

Worked example 4 Adding algebraic fractions

Simplify $\dfrac{3}{x+1} + \dfrac{2}{x-3}$ .

LCD $= (x+1)(x-3)$ .
$\dfrac{3(x-3)}{(x+1)(x-3)} + \dfrac{2(x+1)}{(x+1)(x-3)}$ .
Numerator: $3(x-3) + 2(x+1) = 3x - 9 + 2x + 2 = 5x - 7$ .
Result: $\dfrac{5x - 7}{(x+1)(x-3)}$ .

Worked example 5 Simplifying a quotient of algebraic fractions

Simplify $\dfrac{x^2 - 9}{x + 2} \div \dfrac{x - 3}{x^2 + 2x}$ .

Factorise: $x^2 - 9 = (x-3)(x+3)$ and $x^2 + 2x = x(x+2)$ .
Flip and multiply: $\dfrac{(x-3)(x+3)}{x+2} \times \dfrac{x(x+2)}{x-3}$ .
Cancel $(x-3)$ and $(x+2)$ : $x(x+3)$ or $x^2 + 3x$ .

4. Expanding binomials and factorising quadratics

Formula reference

(a + b)^2 = a^2 + 2ab + b^2 \qquad (a - b)^2 = a^2 - 2ab + b^2

a^2 - b^2 = (a - b)(a + b) \qquad \text{(difference of two squares)}

Worked example 6 Expanding a binomial product

Expand and simplify $(3x - 2)(x + 5)$ .

$3x \cdot x + 3x \cdot 5 + (-2) \cdot x + (-2) \cdot 5$ .
$= 3x^2 + 15x - 2x - 10$ .
$= 3x^2 + 13x - 10$ .

Worked example 7 Factorising using difference of two squares

Factorise $4x^2 - 25$ .

Recognise: $4x^2 = (2x)^2$ and $25 = 5^2$ .
Apply $a^2 - b^2 = (a-b)(a+b)$ : $(2x - 5)(2x + 5)$ .

Worked example 8 Factorising by grouping

Factorise $x^2 + 5x + 2x + 10$ .

Group: $(x^2 + 5x) + (2x + 10)$ .
Factor each group: $x(x + 5) + 2(x + 5)$ .
Common binomial factor: $(x + 5)(x + 2)$ .

Worked example 9 Factorising a non-monic quadratic

Factorise $6x^2 + 11x - 10$ .

Product $= 6 \times (-10) = -60$ . Find two numbers that multiply to $-60$ and add to $11$ : $15$ and $-4$ .
Split: $6x^2 + 15x - 4x - 10$ .
Group: $3x(2x + 5) - 2(2x + 5)$ .
Result: $(2x + 5)(3x - 2)$ .

Worked example 10 Completing the square

Write $x^2 + 8x + 3$ in the form $(x + p)^2 + q$ .

Half the coefficient of $x$ : $p = \dfrac{8}{2} = 4$ .
$(x + 4)^2 = x^2 + 8x + 16$ .
Adjust: $x^2 + 8x + 3 = (x + 4)^2 - 16 + 3 = (x + 4)^2 - 13$ .

5. Substitution and rearranging formulas

To rearrange a formula, use inverse operations to isolate the desired variable — treat every other letter as if it were a number.

Worked example 11 Rearranging a formula

Make $r$ the subject of $V = \dfrac{4}{3}\pi r^3$ .

Multiply both sides by $3$ : $3V = 4\pi r^3$ .
Divide by $4\pi$ : $r^3 = \dfrac{3V}{4\pi}$ .
Cube root: $r = \sqrt[3]{\dfrac{3V}{4\pi}}$ .

Worked example 12 Substituting into a rearranged formula

The formula for the area of a trapezium is $A = \dfrac{1}{2}(a + b)h$ . Find $b$ when $A = 30$ , $a = 4$ and $h = 5$ .

Rearrange: $2A = (a + b)h$ , so $a + b = \dfrac{2A}{h}$ , hence $b = \dfrac{2A}{h} - a$ .
Substitute: $b = \dfrac{2 \times 30}{5} - 4 = 12 - 4 = 8$ .

Practice

Fluency

Tier 1: core skills

Factorise $8x^2 - 12x$ .
Factorise $5a^2b + 10ab^2 - 15ab$ .
Simplify $\dfrac{18x^4 y^2}{6x^2 y^5}$ .
Simplify $(3a^2)^3 \times 2a^{-4}$ .
Expand $(x + 4)(x - 7)$ .
Expand $(2x - 3)^2$ .
Factorise $x^2 - 49$ .
Factorise $x^2 + 3x - 18$ .
Simplify $\dfrac{4}{x} + \dfrac{3}{2x}$ .
Make $t$ the subject of $s = ut + \dfrac{1}{2}at^2$ when $u = 0$ .

Reasoning

Tier 2: multi-step problems

Factorise $3x^2 + 10x - 8$ using grouping.
Write $x^2 - 6x + 1$ in the form $(x - p)^2 + q$ .
Simplify $\dfrac{x^2 - 4}{x^2 + 5x + 6}$ .
Simplify $\dfrac{2}{x - 1} - \dfrac{3}{x + 2}$ .
Make $v$ the subject of $E = \dfrac{1}{2}mv^2$ .
Make $x$ the subject of $y = \dfrac{3x + 1}{x - 2}$ .
Simplify $\dfrac{x^2 + 6x + 9}{x^2 - 9}$ .
Factorise $2x^2 - 18$ completely.

Reasoning

Tier 3: explain and extend

Explain why $x^2 + 4$ cannot be factorised over the real numbers but $x^2 - 4$ can.
By completing the square, show that $x^2 + 6x + 11 > 0$ for all real $x$ .
Simplify $\dfrac{1}{x+1} + \dfrac{1}{x+2} + \dfrac{1}{x+3}$ as a single fraction.
The surface area of a cylinder is $S = 2\pi r^2 + 2\pi rh$ . Factorise the right-hand side, then rearrange for $h$ .
Factorise $x^4 - 16$ completely.

Challenge

Reasoning

Harder reasoning

Factorise $6x^2 - 7xy - 3y^2$ and verify by expanding.
If $a + \dfrac{1}{a} = 5$ , find the value of $a^2 + \dfrac{1}{a^2}$ .
Show that $\dfrac{1}{n} - \dfrac{1}{n+1} = \dfrac{1}{n(n+1)}$ and hence find $\dfrac{1}{1 \times 2} + \dfrac{1}{2 \times 3} + \dfrac{1}{3 \times 4} + \cdots + \dfrac{1}{99 \times 100}$ .
A rectangle has area $x^2 + 7x + 10$ and length $x + 5$ . Find the width, perimeter (in terms of $x$ ), and the value of $x$ if the perimeter is $26$ .

Answers

Answer key

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Year 10 core - answers

Fluency

Tier 1: core skills

$4x(2x - 3)$ .
$5ab(a + 2b - 3)$ .
$\dfrac{3x^2}{y^3}$ .
$(3a^2)^3 = 27a^6$ . Then $27a^6 \times 2a^{-4} = 54a^2$ .
$x^2 - 3x - 28$ .
$4x^2 - 12x + 9$ .
$(x - 7)(x + 7)$ .
$(x + 6)(x - 3)$ .
LCD $= 2x$ : $\dfrac{8}{2x} + \dfrac{3}{2x} = \dfrac{11}{2x}$ .
When $u = 0$ : $s = \dfrac{1}{2}at^2$ , so $t^2 = \dfrac{2s}{a}$ , $t = \sqrt{\dfrac{2s}{a}}$ (taking the positive root).

Reasoning

Tier 2: multi-step problems

Product $= 3 \times (-8) = -24$ . Numbers: $12$ and $-2$ . Split: $3x^2 + 12x - 2x - 8 = 3x(x + 4) - 2(x + 4) = (x + 4)(3x - 2)$ .
$x^2 - 6x + 1 = (x - 3)^2 - 9 + 1 = (x - 3)^2 - 8$ .
$\dfrac{(x-2)(x+2)}{(x+2)(x+3)} = \dfrac{x - 2}{x + 3}$ (for $x \neq -2$ ).
LCD $= (x-1)(x+2)$ : $\dfrac{2(x+2) - 3(x-1)}{(x-1)(x+2)} = \dfrac{2x + 4 - 3x + 3}{(x-1)(x+2)} = \dfrac{-x + 7}{(x-1)(x+2)}$ .
$2E = mv^2$ , $v^2 = \dfrac{2E}{m}$ , $v = \sqrt{\dfrac{2E}{m}}$ .
$y(x - 2) = 3x + 1$ , $xy - 2y = 3x + 1$ , $xy - 3x = 2y + 1$ , $x(y - 3) = 2y + 1$ , $x = \dfrac{2y + 1}{y - 3}$ .
$\dfrac{(x+3)^2}{(x-3)(x+3)} = \dfrac{x + 3}{x - 3}$ (for $x \neq -3$ ).
$2(x^2 - 9) = 2(x - 3)(x + 3)$ .

Reasoning

Tier 3: explain and extend

$x^2 - 4 = (x-2)(x+2)$ uses the difference of two squares. For $x^2 + 4$ , there are no real numbers $a, b$ with $a^2 - b^2 = x^2 + 4$ in the required form. Since $x^2 \geq 0$ , $x^2 + 4 \geq 4 > 0$ , so it never equals zero and cannot be split into real linear factors.
$x^2 + 6x + 11 = (x + 3)^2 - 9 + 11 = (x + 3)^2 + 2$ . Since $(x+3)^2 \geq 0$ , the expression $\geq 2 > 0$ for all real $x$ .
LCD $= (x+1)(x+2)(x+3)$ . Numerator: $(x+2)(x+3) + (x+1)(x+3) + (x+1)(x+2)$ $= (x^2 + 5x + 6) + (x^2 + 4x + 3) + (x^2 + 3x + 2)$ $= 3x^2 + 12x + 11$ . Result: $\dfrac{3x^2 + 12x + 11}{(x+1)(x+2)(x+3)}$ .
$S = 2\pi r(r + h)$ . Rearranging: $r + h = \dfrac{S}{2\pi r}$ , so $h = \dfrac{S}{2\pi r} - r = \dfrac{S - 2\pi r^2}{2\pi r}$ .
$x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$ .

Reasoning

Challenge

Product $= 6 \times (-3) = -18$ . Numbers: $-9$ and $2$ . Split: $6x^2 - 9xy + 2xy - 3y^2 = 3x(2x - 3y) + y(2x - 3y) = (2x - 3y)(3x + y)$ . Check: $(2x - 3y)(3x + y) = 6x^2 + 2xy - 9xy - 3y^2 = 6x^2 - 7xy - 3y^2$ . Correct.
$\left(a + \dfrac{1}{a}\right)^2 = a^2 + 2 + \dfrac{1}{a^2}$ . So $25 = a^2 + 2 + \dfrac{1}{a^2}$ , hence $a^2 + \dfrac{1}{a^2} = 23$ .
$\dfrac{1}{n} - \dfrac{1}{n+1} = \dfrac{(n+1) - n}{n(n+1)} = \dfrac{1}{n(n+1)}$ . The sum telescopes: $\left(\dfrac{1}{1} - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \cdots + \left(\dfrac{1}{99} - \dfrac{1}{100}\right) = 1 - \dfrac{1}{100} = \dfrac{99}{100}$ .
Width $= \dfrac{x^2 + 7x + 10}{x + 5} = \dfrac{(x+5)(x+2)}{x+5} = x + 2$ . Perimeter $= 2(x + 5 + x + 2) = 2(2x + 7) = 4x + 14$ . If perimeter $= 26$ : $4x + 14 = 26$ , $x = 3$ .

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