Level 2 — RecallAlgebra — Introduction & Intermediate

Algebra — Introduction & Intermediate

40 marksprintable — key stays hidden on paper

Level 2 — Recall & Standard Problems

Time: 30 minutes Total Marks: 40

Instructions: Answer all questions. Show working where required. Use $...$ for mathematical expressions.

Q1. [3 marks] Simplify by combining like terms: $5x^2 - 3x + 7 + 2x^2 + 6x - 4$

Q2. [4 marks] Expand using algebraic identities: (a) $(2x + 3)^2$ (b) $(x - 5)(x + 5)$

Q3. [4 marks] Factorise completely: (a) $x^2 - 9x + 20$ (b) $6ab + 3a + 2b + 1$ (by grouping)

Q4. [4 marks] Solve the linear equation by transposition: $3(x - 2) + 4 = 2x + 5$

Q5. [5 marks] The sum of two consecutive even numbers is 46. Form a linear equation and find the numbers.

Q6. [5 marks] Solve the simultaneous equations by elimination: $2x + 3y = 12, \qquad x - y = 1$

Q7. [4 marks] Solve the inequality and represent the solution on a number line: $3x - 5 \le 7$

Q8. [5 marks] For the quadratic equation $x^2 - 7x + 12 = 0$ : (a) Solve by factoring. [3] (b) State the sum and product of the roots and verify using Vieta's formulas. [2]

Q9. [3 marks] Find the remainder when $p(x) = x^3 - 2x^2 + 5x - 3$ is divided by $(x - 2)$ using the remainder theorem.

Q10. [3 marks] Solve the absolute value equation: $|2x - 3| = 7$

END OF PAPER

Answer keyMark scheme & solutions

Q1. [3 marks] Group like terms: $(5x^2 + 2x^2) + (-3x + 6x) + (7 - 4)$ — M1 grouping $= 7x^2 + 3x + 3$ — A2 correct result

Answer: $7x^2 + 3x + 3$

Q2. [4 marks] (a) $(2x+3)^2 = (2x)^2 + 2(2x)(3) + 3^2$ M1 $= 4x^2 + 12x + 9$ A1 (b) $(x-5)(x+5) = x^2 - 5^2$ M1 (difference of squares) $= x^2 - 25$ A1

Q3. [4 marks] (a) Find two numbers multiplying to $20$ , summing to $-9$ : $-4, -5$ . M1 $x^2 - 9x + 20 = (x-4)(x-5)$ A1 (b) Group: $(6ab + 3a) + (2b + 1) = 3a(2b+1) + 1(2b+1)$ M1 $= (2b+1)(3a+1)$ A1

Q4. [4 marks] $3x - 6 + 4 = 2x + 5$ — M1 expand $3x - 2 = 2x + 5$ — A1 simplify $3x - 2x = 5 + 2 \Rightarrow x = 7$ — M1 transpose, A1 answer

Answer: $x = 7$

Q5. [5 marks] Let numbers be $x$ and $x+2$ . M1 Equation: $x + (x+2) = 46$ M1 $2x + 2 = 46 \Rightarrow 2x = 44 \Rightarrow x = 22$ M1, A1 Numbers are $22$ and $24$ . A1

Q6. [5 marks] From eqn 2: multiply by 3: $3x - 3y = 3$ . M1 Add to eqn 1: $2x + 3y = 12$ gives $5x = 15 \Rightarrow x = 3$ M1, A1 Substitute: $3 - y = 1 \Rightarrow y = 2$ M1, A1

Answer: $x = 3, y = 2$

Q7. [4 marks] $3x - 5 \le 7 \Rightarrow 3x \le 12$ M1 $x \le 4$ A1 Number line: closed circle (filled) at $4$ , shading to the left. A2

Q8. [5 marks] (a) $x^2 - 7x + 12 = (x-3)(x-4) = 0$ M1 $x = 3$ or $x = 4$ A2 (b) Sum $= 3 + 4 = 7$ ; Product $= 3 \times 4 = 12$ . M1 Vieta: sum $= -b/a = 7$ , product $= c/a = 12$ ✓ A1

Q9. [3 marks] By remainder theorem, remainder $= p(2)$ . M1 $p(2) = 8 - 8 + 10 - 3 = 7$ M1, A1

Answer: Remainder $= 7$

Q10. [3 marks] $2x - 3 = 7$ or $2x - 3 = -7$ M1 $2x = 10 \Rightarrow x = 5$ ; or $2x = -4 \Rightarrow x = -2$ M1 Answer: $x = 5$ or $x = -2$ A1

[
  {"claim":"Q4: x=7 solves 3(x-2)+4=2x+5","code":"x=symbols('x'); sol=solve(Eq(3*(x-2)+4,2*x+5),x); result=(sol==[7])"},
  {"claim":"Q6: elimination gives x=3,y=2","code":"x,y=symbols('x y'); s=solve([Eq(2*x+3*y,12),Eq(x-y,1)],[x,y]); result=(s[x]==3 and s[y]==2)"},
  {"claim":"Q8: roots of x^2-7x+12 are 3 and 4","code":"x=symbols('x'); r=solve(Eq(x**2-7*x+12,0),x); result=(set(r)=={3,4})"},
  {"claim":"Q9: remainder p(2)=7","code":"x=symbols('x'); p=x**3-2*x**2+5*x-3; result=(p.subs(x,2)==7)"},
  {"claim":"Q10: |2x-3|=7 gives x=5 or -2","code":"x=symbols('x'); r=solve(Eq(Abs(2*x-3),7),x); result=(set(r)=={5,-2})"}
]