How to use this bank.
- Start at Level 1 and move up — don't jump to Level 4 first.
- Factorise fully before doing anything else — most methods on this page start there.
- For inequalities, watch for a negative multiplier or divisor — that's the one place the rule changes.
- Reveal the memo only after a genuine attempt.
L1 — Knowledge 4 Qs
L2 — Routine Procedures 5 Qs
L3 — Complex Procedures 4 Qs
L4 — Problem Solving 4 Qs
24%
Level 1 | Knowledge
Direct Recall
Definitions and one-step applications of a rule already stated.
Exponents
Zero Exponent
Simplify: \((2x)^0\), \(x\neq0\)
Memo
✓ The whole expression \((2x)^0\) is being raised to the power \(0\) — and any nonzero base raised to the power \(0\) equals \(1\), regardless of what that base actually is.✓ Since \(x\neq0\) is given, \(2x\neq0\), so this rule applies directly: \(\boxed{1}\)Real Numbers
Locating a Surd
Between which two consecutive integers does \(\sqrt{45}\) lie?
Memo
✓ To trap \(\sqrt{45}\) between two consecutive integers, find the perfect squares closest to \(45\) on either side: \(36=6^2\) and \(49=7^2\).✓ Since \(36<45<49\), and taking square roots of positive numbers preserves their order, the same relationship holds for the roots themselves: \(\boxed{6<\sqrt{45}<7}\)Factorising
Common Factor
Factorise: \(5x^2-15x\)
Memo
✓ Every term shares a factor of \(5x\): both coefficients (\(5\) and \(15\)) share a factor of \(5\), and both terms contain at least one \(x\).✓ Divide each term by \(5x\) and write it outside a bracket: \(\boxed{5x(x-3)}\) (check by expanding: \(5x(x-3)=5x^2-15x\)✓)Exponents
Negative Exponent
Rewrite \(x^{-3}\) using a positive exponent.
Memo
✓ A negative exponent is an instruction to take the reciprocal — it does not make the value negative: \(x^{-n}=\dfrac{1}{x^n}\).✓ Applying that here: \(\boxed{\dfrac{1}{x^3}}\)29%
Level 2 | Routine Procedures
One Established Method
Factorising, exponents, exponential equations and algebraic fractions, applying one method cleanly.
Factorising
Perfect Square Trinomial
Factorise fully: \(x^2-4x+4\)
Memo
✓ Find two numbers that multiply to the constant \(4\) and add to the middle coefficient \(-4\): both numbers must be \(-2\), since \((-2)\times(-2)=4\) and \((-2)+(-2)=-4\).✓ A repeated factor is written as a square: \(\boxed{(x-2)^2}\) (check: \((x-2)^2=x^2-2x-2x+4=x^2-4x+4\)✓)Linear Equations
Solving a Linear Equation
Solve for \(x\): \(3(x-2)=x+6\)
Memo
✓ Expand the bracket first, multiplying every term inside it by \(3\): \(3x-6=x+6\)✓ Collect the \(x\)-terms on one side and the constants on the other — subtract \(x\) from both sides and add \(6\) to both sides: \(2x=12\)✓ Divide both sides by \(2\): \(\boxed{x=6}\) (check: \(3(6-2)=12\) and \(6+6=12\)✓)Exponents
Power of a Product
Simplify: \((2x^2y)^3\)
Memo
✓ The power outside the bracket applies to EVERY factor inside it, including the plain number \(2\), not only the letters: cube every factor separately: \(2^3\cdot x^{2\times3}\cdot y^3\)✓ Evaluate each piece: \(2^3=8\), \(x^{2\times3}=x^6\), \(y^{1\times3}=y^3\): \(\boxed{8x^6y^3}\)Exponential Equations
Same-Base Exponential Equation
Solve for \(x\): \(5^x=125\)
Memo
✓ Exponents can only be equated once both sides share the SAME base — rewrite \(125\) as a power of \(5\): \(125=5^3\), so \(5^x=5^3\)✓ Now the bases match, so the exponents themselves must be equal: \(\boxed{x=3}\)Algebraic Fractions
Simplify Using a Difference of Squares
Simplify fully: \(\dfrac{x^2-25}{x+5}\)
Memo
✓ The numerator is a difference of two squares (\(x^2\) and \(5^2\)): factorise it as \((x-5)(x+5)\), giving \(\dfrac{(x-5)(x+5)}{x+5}\)✓ The factor \((x+5)\) now appears in both the top and the bottom, so it cancels — the value it excludes (whatever would make the ORIGINAL denominator zero) must still be stated: \(\boxed{x-5,\ x\neq-5}\)24%
Level 3 | Complex Procedures
Multi-Step Methods
Harder trinomials, the sum of two cubes, and simultaneous or quadratic equations.
Sum of Cubes
Factorise a Sum of Two Cubes
Factorise fully: \(x^3+27\)
Memo
✓ Both terms are perfect cubes: \(x^3\) and \(27=3^3\), so \(a=x,\,b=3\). Apply the sum-of-cubes formula \(a^3+b^3=(a+b)(a^2-ab+b^2)\).✓ Substituting gives \(\boxed{(x+3)(x^2-3x+9)}\) — notice the middle term of the second bracket takes the OPPOSITE sign to the sum in the first bracket, exactly as the formula requires.Factorising
Trinomial, Leading Coefficient 3
Factorise fully: \(3x^2-2x-5\)
Memo
✓ Multiply the leading coefficient by the constant to find the target product: \(3\times(-5)=-15\); then find two numbers multiplying to \(-15\) and adding to the middle coefficient, \(-2\): these are \(-5\) and \(3\)✓ Split the middle term using these numbers, then group in pairs and factorise each group: \(3x^2-5x+3x-5=x(3x-5)+1(3x-5)\)✓ Both groups now share the factor \((3x-5)\): \(\boxed{(x+1)(3x-5)}\) (check: \((x+1)(3x-5)=3x^2-5x+3x-5=3x^2-2x-5\)✓)Simultaneous Equations
Simultaneous Linear Equations
Solve simultaneously: \(x+y=10\) and \(2x-y=5\)
Memo
✓ The \(y\)-coefficients, \(+y\) and \(-y\), are already opposites, so adding the two equations directly cancels \(y\): \((x+y)+(2x-y)=10+5\Rightarrow3x=15\Rightarrow x=5\)✓ Substitute \(x=5\) into the simpler equation, \(x+y=10\), to find \(y\): \(y=5\)✓ \(\boxed{x=5,\ y=5}\) — check in the other equation: \(2(5)-5=5\)✓Quadratic Equations
Quadratic Equation, Leading Coefficient 2
Solve for \(x\): \(2x^2+x-3=0\)
Memo
✓ Multiply the leading coefficient by the constant: \(2\times(-3)=-6\); find two numbers multiplying to \(-6\) and adding to the middle coefficient, \(1\): these are \(3\) and \(-2\)✓ Split the middle term and group in pairs: \(2x^2+3x-2x-3=x(2x+3)-1(2x+3)=(x-1)(2x+3)\)✓ Zero-product law — each bracket set to zero in turn: \(\boxed{x=1\text{ or }x=-1{,}5}\)23%
Level 4 | Problem Solving
Combined Skills
Chaining factorisation, exponential equations, inequalities and word problems together.
Algebraic FractionsDifference of Cubes
Simplify With a Cube Denominator
Simplify fully: \(\dfrac{x^2-9}{x^3-27}\)
Memo
✓ Factorise the numerator as a difference of two squares: \(x^2-9=(x-3)(x+3)\)✓ Factorise the denominator as a difference of two cubes, \(x^3-3^3\): \(x^3-27=(x-3)(x^2+3x+9)\)✓ Both top and bottom share the factor \((x-3)\), so cancel it — the excluded value is whatever made the ORIGINAL denominator zero: \(\boxed{\dfrac{x+3}{x^2+3x+9},\ x\neq3}\)Exponential Equations
Different Starting Bases
Solve for \(x\): \(4^{x+1}=2^{3x-1}\)
Memo
✓ The two sides start from different bases (\(4\) and \(2\)), so they can't be equated yet — rewrite \(4\) as a power of \(2\): \(4^{x+1}=(2^2)^{x+1}=2^{2x+2}\)✓ Both sides now share base \(2\), so the exponents themselves must be equal: \(2x+2=3x-1\)✓ Solve for \(x\): \(\boxed{x=3}\)Inequalities
Double Inequality
Solve for \(x\), and write your answer using interval notation: \(-2\le3-x<7\)
Memo
✓ Subtract \(3\) from all three parts to start isolating \(x\) — subtraction never flips an inequality: \(-5\le-x<4\)✓ Divide all three parts by \(-1\): since this is a negative number, BOTH inequality signs must flip: \(5\ge x>-4\)✓ Rewrite from smallest to largest so it reads naturally: \(-4Word ProblemQuadratic Equations
Two Numbers, a Difference and a Product
Two positive numbers differ by 4, and their product is 45. Determine the two numbers.
Memo
✓ Let the smaller number be \(x\); "differ by 4" means the larger number is \(x+4\)✓ Translate "their product is 45" directly into an equation, then move everything to one side to solve: \(x(x+4)=45\Rightarrow x^2+4x-45=0\)✓ Factorise: \((x+9)(x-5)=0\Rightarrow x=-9\) or \(x=5\)✓ The problem specifies positive numbers, so reject the negative solution \(x=-9\): \(\boxed{\text{the numbers are }5\text{ and }9}\) (check: \(5\times9=45\)✓, \(9-5=4\)✓)