Year 10 Mathematics | Victorian Curriculum 2.0
Algebraic techniques
Topic 02 | Number & Algebra | Practice

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.
Why master algebraic techniques?

Algebra is the language of generalisation. Being able to factorise, expand and simplify lets you solve equations, model real situations and prove results. These techniques are essential prerequisites for every branch of senior mathematics — from calculus to statistics.

Where you'll see this
  • Physics: rearranging v=u+atv = u + atv=u+at or E=mc2E = mc^2E=mc2 to solve for any variable.
  • Economics: simplifying cost/revenue expressions to find break-even points.
  • Engineering: factorising to find when structural loads equal zero.
  • Medicine: rearranging dosage formulas for patient weight.
Worked example 0 Real-world example: rearranging a physics formula

The kinetic energy formula is E=12mv2E = \dfrac{1}{2}mv^2E=21​mv2. A vehicle of mass 120012001200 kg has kinetic energy 48 00048\,00048000 J. Find its speed.

  1. Rearrange for vvv: multiply both sides by 222: 2E=mv22E = mv^22E=mv2.
  2. Divide by mmm: v2=2Emv^2 = \dfrac{2E}{m}v2=m2E​.
  3. Take the square root: v=2Emv = \sqrt{\dfrac{2E}{m}}v=m2E​​.
  4. Substitute: v=2×48 0001200=80=45≈8.94v = \sqrt{\dfrac{2 \times 48\,000}{1200}} = \sqrt{80} = 4\sqrt{5} \approx 8.94v=12002×48000​​=80​=45​≈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)ab + ac = a(b + c)ab+ac=a(b+c)
Worked example 1 Taking out a common factor

Factorise 6x3y−9x2y2+3xy6x^3y - 9x^2y^2 + 3xy6x3y−9x2y2+3xy.

  1. Numerical HCF: gcd⁡(6,9,3)=3\gcd(6, 9, 3) = 3gcd(6,9,3)=3.
  2. Variable HCF: xxx (lowest power) and yyy (lowest power) =xy= xy=xy.
  3. HCF =3xy= 3xy=3xy. Divide each term: 6x3y3xy=2x2\dfrac{6x^3y}{3xy} = 2x^23xy6x3y​=2x2, −9x2y23xy=−3xy\dfrac{-9x^2y^2}{3xy} = -3xy3xy−9x2y2​=−3xy, 3xy3xy=1\dfrac{3xy}{3xy} = 13xy3xy​=1.
  4. Result: 3xy(2x2−3xy+1)3xy(2x^2 - 3xy + 1)3xy(2x2−3xy+1).

2. Exponent laws for products and quotients

Formula reference

am×an=am+naman=am−n(am)n=amna^m \times a^n = a^{m+n} \qquad \frac{a^m}{a^n} = a^{m-n} \qquad (a^m)^n = a^{mn}am×an=am+nanam​=am−n(am)n=amn(ab)n=anbn(ab)n=anbna0=1  (a≠0)(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)(ab)n=anbn(ba​)n=bnan​a0=1(a=0)
Worked example 2 Simplifying with exponent laws

Simplify 12a5b34a2b7\dfrac{12a^5 b^3}{4a^2 b^7}4a2b712a5b3​.

  1. Coefficients: 124=3\dfrac{12}{4} = 3412​=3.
  2. aaa terms: a5−2=a3a^{5-2} = a^3a5−2=a3.
  3. bbb terms: b3−7=b−4=1b4b^{3-7} = b^{-4} = \dfrac{1}{b^4}b3−7=b−4=b41​.
  4. Result: 3a3b4\dfrac{3a^3}{b^4}b43a3​.
Worked example 3 Simplifying a product of powers

Simplify (2x3)4×3x−2(2x^3)^4 \times 3x^{-2}(2x3)4×3x−2.

  1. (2x3)4=24⋅x12=16x12(2x^3)^4 = 2^4 \cdot x^{12} = 16x^{12}(2x3)4=24⋅x12=16x12.
  2. 16x12×3x−2=48x1016x^{12} \times 3x^{-2} = 48x^{10}16x12×3x−2=48x10.

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 3x+1+2x−3\dfrac{3}{x+1} + \dfrac{2}{x-3}x+13​+x−32​.

  1. LCD =(x+1)(x−3)= (x+1)(x-3)=(x+1)(x−3).
  2. 3(x−3)(x+1)(x−3)+2(x+1)(x+1)(x−3)\dfrac{3(x-3)}{(x+1)(x-3)} + \dfrac{2(x+1)}{(x+1)(x-3)}(x+1)(x−3)3(x−3)​+(x+1)(x−3)2(x+1)​.
  3. Numerator: 3(x−3)+2(x+1)=3x−9+2x+2=5x−73(x-3) + 2(x+1) = 3x - 9 + 2x + 2 = 5x - 73(x−3)+2(x+1)=3x−9+2x+2=5x−7.
  4. Result: 5x−7(x+1)(x−3)\dfrac{5x - 7}{(x+1)(x-3)}(x+1)(x−3)5x−7​.
Worked example 5 Simplifying a quotient of algebraic fractions

Simplify x2−9x+2÷x−3x2+2x\dfrac{x^2 - 9}{x + 2} \div \dfrac{x - 3}{x^2 + 2x}x+2x2−9​÷x2+2xx−3​.

  1. Factorise: x2−9=(x−3)(x+3)x^2 - 9 = (x-3)(x+3)x2−9=(x−3)(x+3) and x2+2x=x(x+2)x^2 + 2x = x(x+2)x2+2x=x(x+2).
  2. Flip and multiply: (x−3)(x+3)x+2×x(x+2)x−3\dfrac{(x-3)(x+3)}{x+2} \times \dfrac{x(x+2)}{x-3}x+2(x−3)(x+3)​×x−3x(x+2)​.
  3. Cancel (x−3)(x-3)(x−3) and (x+2)(x+2)(x+2): x(x+3)x(x+3)x(x+3) or x2+3xx^2 + 3xx2+3x.

4. Expanding binomials and factorising quadratics

Formula reference

(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2(a + b)^2 = a^2 + 2ab + b^2 \qquad (a - b)^2 = a^2 - 2ab + b^2(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2a2−b2=(a−b)(a+b)(difference of two squares)a^2 - b^2 = (a - b)(a + b) \qquad \text{(difference of two squares)}a2−b2=(a−b)(a+b)(difference of two squares)
Worked example 6 Expanding a binomial product

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

  1. 3x⋅x+3x⋅5+(−2)⋅x+(−2)⋅53x \cdot x + 3x \cdot 5 + (-2) \cdot x + (-2) \cdot 53x⋅x+3x⋅5+(−2)⋅x+(−2)⋅5.
  2. =3x2+15x−2x−10= 3x^2 + 15x - 2x - 10=3x2+15x−2x−10.
  3. =3x2+13x−10= 3x^2 + 13x - 10=3x2+13x−10.
Worked example 7 Factorising using difference of two squares

Factorise 4x2−254x^2 - 254x2−25.

  1. Recognise: 4x2=(2x)24x^2 = (2x)^24x2=(2x)2 and 25=5225 = 5^225=52.
  2. Apply a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b): (2x−5)(2x+5)(2x - 5)(2x + 5)(2x−5)(2x+5).
Worked example 8 Factorising by grouping

Factorise x2+5x+2x+10x^2 + 5x + 2x + 10x2+5x+2x+10.

  1. Group: (x2+5x)+(2x+10)(x^2 + 5x) + (2x + 10)(x2+5x)+(2x+10).
  2. Factor each group: x(x+5)+2(x+5)x(x + 5) + 2(x + 5)x(x+5)+2(x+5).
  3. Common binomial factor: (x+5)(x+2)(x + 5)(x + 2)(x+5)(x+2).
Worked example 9 Factorising a non-monic quadratic

Factorise 6x2+11x−106x^2 + 11x - 106x2+11x−10.

  1. Product =6×(−10)=−60= 6 \times (-10) = -60=6×(−10)=−60. Find two numbers that multiply to −60-60−60 and add to 111111: 151515 and −4-4−4.
  2. Split: 6x2+15x−4x−106x^2 + 15x - 4x - 106x2+15x−4x−10.
  3. Group: 3x(2x+5)−2(2x+5)3x(2x + 5) - 2(2x + 5)3x(2x+5)−2(2x+5).
  4. Result: (2x+5)(3x−2)(2x + 5)(3x - 2)(2x+5)(3x−2).
Worked example 10 Completing the square

Write x2+8x+3x^2 + 8x + 3x2+8x+3 in the form (x+p)2+q(x + p)^2 + q(x+p)2+q.

  1. Half the coefficient of xxx: p=82=4p = \dfrac{8}{2} = 4p=28​=4.
  2. (x+4)2=x2+8x+16(x + 4)^2 = x^2 + 8x + 16(x+4)2=x2+8x+16.
  3. Adjust: x2+8x+3=(x+4)2−16+3=(x+4)2−13x^2 + 8x + 3 = (x + 4)^2 - 16 + 3 = (x + 4)^2 - 13x2+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 rrr the subject of V=43πr3V = \dfrac{4}{3}\pi r^3V=34​πr3.

  1. Multiply both sides by 333: 3V=4πr33V = 4\pi r^33V=4πr3.
  2. Divide by 4π4\pi4π: r3=3V4πr^3 = \dfrac{3V}{4\pi}r3=4π3V​.
  3. Cube root: r=3V4π3r = \sqrt[3]{\dfrac{3V}{4\pi}}r=34π3V​​.
Worked example 12 Substituting into a rearranged formula

The formula for the area of a trapezium is A=12(a+b)hA = \dfrac{1}{2}(a + b)hA=21​(a+b)h. Find bbb when A=30A = 30A=30, a=4a = 4a=4 and h=5h = 5h=5.

  1. Rearrange: 2A=(a+b)h2A = (a + b)h2A=(a+b)h, so a+b=2Aha + b = \dfrac{2A}{h}a+b=h2A​, hence b=2Ah−ab = \dfrac{2A}{h} - ab=h2A​−a.
  2. Substitute: b=2×305−4=12−4=8b = \dfrac{2 \times 30}{5} - 4 = 12 - 4 = 8b=52×30​−4=12−4=8.

Practice

Fluency

Tier 1: core skills

    1. Factorise 8x2−12x8x^2 - 12x8x2−12x.
    2. Factorise 5a2b+10ab2−15ab5a^2b + 10ab^2 - 15ab5a2b+10ab2−15ab.
    3. Simplify 18x4y26x2y5\dfrac{18x^4 y^2}{6x^2 y^5}6x2y518x4y2​.
    4. Simplify (3a2)3×2a−4(3a^2)^3 \times 2a^{-4}(3a2)3×2a−4.
    5. Expand (x+4)(x−7)(x + 4)(x - 7)(x+4)(x−7).
    6. Expand (2x−3)2(2x - 3)^2(2x−3)2.
    7. Factorise x2−49x^2 - 49x2−49.
    8. Factorise x2+3x−18x^2 + 3x - 18x2+3x−18.
    9. Simplify 4x+32x\dfrac{4}{x} + \dfrac{3}{2x}x4​+2x3​.
    10. Make ttt the subject of s=ut+12at2s = ut + \dfrac{1}{2}at^2s=ut+21​at2 when u=0u = 0u=0.
Reasoning

Tier 2: multi-step problems

    1. Factorise 3x2+10x−83x^2 + 10x - 83x2+10x−8 using grouping.
    2. Write x2−6x+1x^2 - 6x + 1x2−6x+1 in the form (x−p)2+q(x - p)^2 + q(x−p)2+q.
    3. Simplify x2−4x2+5x+6\dfrac{x^2 - 4}{x^2 + 5x + 6}x2+5x+6x2−4​.
    4. Simplify 2x−1−3x+2\dfrac{2}{x - 1} - \dfrac{3}{x + 2}x−12​−x+23​.
    5. Make vvv the subject of E=12mv2E = \dfrac{1}{2}mv^2E=21​mv2.
    6. Make xxx the subject of y=3x+1x−2y = \dfrac{3x + 1}{x - 2}y=x−23x+1​.
    7. Simplify x2+6x+9x2−9\dfrac{x^2 + 6x + 9}{x^2 - 9}x2−9x2+6x+9​.
    8. Factorise 2x2−182x^2 - 182x2−18 completely.
Reasoning

Tier 3: explain and extend

    1. Explain why x2+4x^2 + 4x2+4 cannot be factorised over the real numbers but x2−4x^2 - 4x2−4 can.
    2. By completing the square, show that x2+6x+11>0x^2 + 6x + 11 > 0x2+6x+11>0 for all real xxx.
    3. Simplify 1x+1+1x+2+1x+3\dfrac{1}{x+1} + \dfrac{1}{x+2} + \dfrac{1}{x+3}x+11​+x+21​+x+31​ as a single fraction.
    4. The surface area of a cylinder is S=2πr2+2πrhS = 2\pi r^2 + 2\pi rhS=2πr2+2πrh. Factorise the right-hand side, then rearrange for hhh.
    5. Factorise x4−16x^4 - 16x4−16 completely.

Challenge

Reasoning

Harder reasoning

    1. Factorise 6x2−7xy−3y26x^2 - 7xy - 3y^26x2−7xy−3y2 and verify by expanding.
    2. If a+1a=5a + \dfrac{1}{a} = 5a+a1​=5, find the value of a2+1a2a^2 + \dfrac{1}{a^2}a2+a21​.
    3. Show that 1n−1n+1=1n(n+1)\dfrac{1}{n} - \dfrac{1}{n+1} = \dfrac{1}{n(n+1)}n1​−n+11​=n(n+1)1​ and hence find 11×2+12×3+13×4+⋯+199×100\dfrac{1}{1 \times 2} + \dfrac{1}{2 \times 3} + \dfrac{1}{3 \times 4} + \cdots + \dfrac{1}{99 \times 100}1×21​+2×31​+3×41​+⋯+99×1001​.
    4. A rectangle has area x2+7x+10x^2 + 7x + 10x2+7x+10 and length x+5x + 5x+5. Find the width, perimeter (in terms of xxx), and the value of xxx if the perimeter is 262626.
Year 10 Mathematics study companion | Practice