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Optional Week 0 · Foundation Diagnostic

Check the foundations before the intensive

Try all 18 questions here. Show your working, then open each solution to see what needs repair. This is a 60-mark starting check; allow about 70 minutes.

Three ideas to check first

Keep index laws tied to the operation

For the same nonzero base, multiplying powers adds indices: x⁻² × x⁵ = x³. Subtracting powers as terms needs a common denominator instead.

Factor the whole expression

9a² − 16b² = (3a−4b)(3a+4b). Check by expanding both factors; the cross terms cancel.

Write restrictions before simplifying

For (x²−4)(x²+3x+2), factor to (x−2)(x+2)[(x+1)(x+2)]. The result is (x−2)(x+1), while x=−1,−2 remain excluded from the original expression.

Foundation practice

Diagnostic foundations

Check three short prerequisites before moving into longer work.

Question 1

2 marks

Simplify (3x⁴y⁻²)(2x⁻¹y⁵) using positive indices. State restrictions.

See the worked solution
  1. Use the product law separately for x and y.
  2. Coefficient: 3×2=6.
  3. x exponent: 4+(−1)=3; y exponent: −2+5=3.
  4. Answer: 6x³y³. Negative exponents in the original require x≠0 and y≠0.

Question 2

2 marks

Rewrite 823 as an integer. Explain the denominator of the index.

See the worked solution
  1. 23 means cube root, then square (or square then cube root).
  2. ∛8=2, so 823=2²=4.

Question 3

2 marks

Factorise completely 12u² − 27v².

See the worked solution
  1. Take out 3: 3(4u²−9v²).
  2. Difference of squares: 4u²−9v²=(2u−3v)(2u+3v).
  3. Answer: 3(2u−3v)(2u+3v).

Foundation practice

Core repair

Use the right operation and keep restrictions from the original expression.

Question 4

4 marks

Simplify (15a³b⁻²)(8c) ÷ (25a⁻¹b)(12c²) using positive indices. State restrictions.

See the worked solution
  1. Change division to multiplication by the reciprocal of the second fraction.
  2. (15a³b⁻²)(8c) × (12c²)(25a⁻¹b).
  3. Coefficients: 180200=910.
  4. a exponent 3−(−1)=4; b exponent −2−1=−3; c exponent −1+2=1.
  5. Answer: 9a⁴c(10b³). Restrictions: a,b,c≠0.

Question 5

3 marks

Given 2x+1 × 4x−2 = 32, solve for x. Show common bases and check.

See the worked solution
  1. Rewrite all terms with base 2: 4=2² and 32=2⁵.
  2. 2x+1×22x−4=2⁵, hence 3x−3=5.
  3. Add 3: 3x=8, so x=83.
  4. Check: at x=83, (x+1)+2(x−2)=113+43=5, so the left side is 2⁵=32.

Question 6

3 marks

Rectangle sides are 3x²y and 2xy⁻¹, with x>0,y>0. Find area and perimeter. Explain why perimeter terms cannot combine.

See the worked solution
  1. Area = (3x²y)(2xy⁻¹)=6x³.
  2. Perimeter =2(3x²y+2xy⁻¹)=6x²y+4xy⁻¹.
  3. x,y>0 ensures both side lengths are positive. The terms have different powers of x and y, so they are unlike terms and cannot combine.

Question 7

3 marks

For positive m,n, factorise 18m³n − 8mn³ completely. Hence solve 18m³n − 8mn³ = 0 for n in terms of m.

See the worked solution
  1. Factor 2mn: 2mn(9m²−4n²).
  2. Apply difference of squares: 2mn(3m−2n)(3m+2n).
  3. For positive m,n, 2mn and 3m+2n are nonzero. Thus 3m−2n=0, so n=3m2.

Question 8

3 marks

Simplify 3(2x−1) − 7(x+1) as one fraction. State excluded values.

See the worked solution
  1. Common denominator: (2x−1)(x+1).
  2. Numerator: 3(x+1)−7(2x−1)=3x+3−14x+7=10−11x.
  3. Answer: (10−11x)((2x−1)(x+1)).
  4. Original denominators require x≠12 and x≠−1.

Foundation practice

Multistep practice

Connect index laws, algebra and prime factors, then check what an answer means.

Question 9

4 marks

Simplify [(12x²y⁻¹)(5z)] ÷ [(18x⁻¹y²)(25z²)] using positive indices. State restrictions, then evaluate at x=−1,y=2,z=3.

See the worked solution
  1. Invert only the divisor: first fraction × (25z²)(18x⁻¹y²).
  2. Coefficient: (12×25)(5×18)=103.
  3. x exponent 2−(−1)=3; y exponent −1−2=−3; z exponent −1+2=1.
  4. Answer: 10x³z(3y³), with x,y,z≠0.
  5. At (−1,2,3): 10(−1)³(3)(3·2³)=−3024=−54.

Question 10

4 marks

Solve 32x−1 = 27x−2 × 9². Explain why common-base comparison is reversible.

See the worked solution
  1. Use 27=3³ and 9²=3⁴. RHS becomes 33(x−2)+4=33x−2.
  2. LHS is 32x−1. Since base 3 is positive and not 1, equal powers imply equal exponents.
  3. 2x−1=3x−2, so x=1.
  4. Check: LHS=3; RHS=27⁻¹×81=3.

Question 11

4 marks

A=2ʳ3ˢ and B=2ᵗ3ᵘ, with r,s,t,u non-negative integers and AB an integer. Give necessary and sufficient conditions for AB to be (i) a positive integer square, (ii) a positive integer cube. Explain.

See the worked solution
  1. AB=2r−t3s−u. Since AB is an integer, both differences are non-negative integers.
  2. It is a positive integer square iff both differences are even.
  3. It is a positive integer cube iff both differences are multiples of 3.
  4. Non-negativity is needed for integer quotient; parity/multiples ensure square/cube exponents.

Question 12

4 marks

A=(x+2)²−(x−2)². Simplify by expansion and by an identity; explain cancellation, then find x when A=40.

See the worked solution
  1. Expansion: (x²+4x+4)−(x²−4x+4)=8x.
  2. Difference of squares: [(x+2)−(x−2)][(x+2)+(x−2)]=(4)(2x)=8x.
  3. The x² and constant terms cancel; the cross terms remain.
  4. Set 8x=40, giving x=5.

Question 13

4 marks

N=2³×3²×5. Find least positive k,j so Nk is a perfect square and Nj a perfect cube. Explain exponent adjustments.

See the worked solution
  1. For a square, every prime exponent must be even. 2³ needs one extra 2; 3² is already even; 5¹ needs one extra 5.
  2. Least k=2×5=10, giving exponents 4,2,2.
  3. For a cube, exponents must be multiples of 3: 2³ is already suitable; add one 3 and 5², so j=3×25=75.
  4. Answer: k=10 and j=75.

Foundation practice

Spot the mistake

Explain why a student's step fails before correcting it.

Question 14

3 marks

A student writes x⁻²−x⁻³=x. Find the correct simplification and domain. Identify the mistaken exponent operation.

See the worked solution
  1. Write with common denominator x³: 1x²−1x³=(x−1)x³.
  2. Domain: x≠0, from negative powers.
  3. The exponent subtraction law applies to division of like bases, not subtraction.

Question 15

3 marks

A student expands (2r−3)² as 4r²+9. Correct it and explain the missing term.

See the worked solution
  1. (2r−3)²=(2r)²−2(2r)(3)+3².
  2. So the result is 4r²−12r+9. The middle term is twice the product of the two terms, with the sign from −3.

Question 16

3 marks

A student simplifies x(x+4)(x+4) to x+4. Give correct simplification and restriction. Explain what factor is cancelled.

See the worked solution
  1. Factor numerator and denominator as x·(x+4)(x+4).
  2. For x≠−4, cancel the common factor x+4.
  3. Answer: x, with original restriction x≠−4. Cancellation does not leave x+4 behind.

Foundation practice

Optional stretch

Apply familiar facts in a less familiar structure. These two questions are optional after the main repair work.

Question 17

5 marks

Find the least positive n such that 72n is both a perfect square and a perfect cube. Justify the prime-exponent conditions.

See the worked solution
  1. 72=2³×3². For a number to be both a square and cube, each prime exponent must be a multiple of lcm(2,3)=6.
  2. Raise the 2 exponent from 3 to 6: multiply by 2³. Raise 3 from 2 to 6: multiply by 3⁴.
  3. Least n=2³×3⁴=8×81=648. Then 72n=2⁶×3⁶=6⁶, which is both a square and cube.

Question 18

4 marks

Solve 2x+1=82−x. Explain why common-base comparison works and whether any real x is excluded.

See the worked solution
  1. Rewrite 8 as 2³: RHS=23(2−x)=26−3x.
  2. Equal powers of base 2 give x+1=6−3x; hence 4x=5, x=54.
  3. Both sides are defined for every real x; no values are excluded.

Prefer paper?

The same 18 questions and solutions are available to print.

Worksheet PDF ↓Worked solutions PDF ↓

What to practise next

Week 1 · Intensive AlgebraContinue into the intensive Indices guideRepair a specific prerequisite Factorisation guideReview the method and a harder question