Level 1 — RecognitionAlgebra — Introduction & Intermediate

Algebra — Introduction & Intermediate

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Level 1 — Recognition

Time limit: 20 minutes Total marks: 30

Section A — Multiple Choice (1 mark each)

Choose the single best option.

Q1. In the expression $7x^2 - 3x + 5$ , the coefficient of $x$ is: (a) $7$ (b) $-3$ (c) $3$ (d) $5$

Q2. Which of the following are like terms? (a) $3x^2,\ 3x$ (b) $5xy,\ -2yx$ (c) $4a,\ 4b$ (d) $x^2y,\ xy^2$

Q3. The identity $(a+b)(a-b)$ equals: (a) $a^2+b^2$ (b) $a^2-2ab+b^2$ (c) $a^2-b^2$ (d) $a^2+2ab+b^2$

Q4. The degree of the polynomial $4x^3 - 2x^5 + x$ is: (a) $3$ (b) $5$ (c) $1$ (d) $9$

Q5. A polynomial with exactly three terms is called a: (a) monomial (b) binomial (c) trinomial (d) constant

Q6. The solution of $2x + 6 = 0$ is: (a) $x=3$ (b) $x=-3$ (c) $x=6$ (d) $x=-6$

Q7. For the quadratic $ax^2+bx+c=0$ , the discriminant is: (a) $b^2-4ac$ (b) $b^2+4ac$ (c) $\sqrt{b^2-4ac}$ (d) $-b/2a$

Q8. If the discriminant $D>0$ , the roots are: (a) real and equal (b) real and distinct (c) complex (d) zero

Q9. By Vieta's formulas, the sum of the roots of $x^2 - 5x + 6 = 0$ is: (a) $6$ (b) $-5$ (c) $5$ (d) $-6$

Q10. Rationalizing $\dfrac{1}{\sqrt{3}}$ gives: (a) $\dfrac{\sqrt{3}}{3}$ (b) $\sqrt{3}$ (c) $3\sqrt{3}$ (d) $\dfrac{1}{3}$

Section B — Matching (1 mark each pair; 5 marks)

Q11. Match Column A with Column B.

Column A	Column B
(i) $(a+b)^2$	(P) $a^3+b^3$
(ii) $(a-b)^2$	(Q) $a^2+2ab+b^2$
(iii) $(a+b)(a^2-ab+b^2)$	(R) $a^2-2ab+b^2$
(iv) $(a-b)(a^2+ab+b^2)$	(S) $a^3-b^3$
(v) $(a+b)(a-b)$	(T) $a^2-b^2$

Section C — True / False with justification (2 marks each; 15 marks)

State True or False and give a one-line reason.

Q12. $x=2$ is a root of $x^2 - 4 = 0$ .

Q13. The terms $5ab$ and $5ba$ are unlike terms.

Q14. $(x+3)$ is a factor of $x^2 + 5x + 6$ .

Q15. The inequality $-2x > 6$ has solution $x > -3$ .

Q16. $\sqrt{16} + \sqrt{9} = \sqrt{25}$ .

Q17. By the Remainder Theorem, dividing $p(x)=x^2+x+1$ by $(x-1)$ leaves remainder $3$ .

Q18. The equation $|x| = -4$ has no solution.

Answer keyMark scheme & solutions

Section A

Q1. (b) $-3$ . The coefficient of $x$ is the number multiplying $x$ . (1)

Q2. (b) $5xy$ and $-2yx$ . Since $yx=xy$ , both have identical variable part $xy$ ; like terms need the same variables with same powers. (1)

Q3. (c) $a^2-b^2$ . Difference of squares identity. (1)

Q4. (b) $5$ . Highest power of $x$ is $5$ (from $-2x^5$ ). (1)

Q5. (c) trinomial. Three terms → trinomial. (1)

Q6. (b) $x=-3$ . $2x=-6\Rightarrow x=-3$ . (1)

Q7. (a) $b^2-4ac$ . Definition of discriminant. (1)

Q8. (b) real and distinct. $D>0$ ⇒ two unequal real roots. (1)

Q9. (c) $5$ . Sum $=-b/a = -(-5)/1 = 5$ . (1)

Q10. (a) $\dfrac{\sqrt3}{3}$ . Multiply top and bottom by $\sqrt3$ : $\dfrac{\sqrt3}{3}$ . (1)

Section B

Q11. (i)→Q, (ii)→R, (iii)→P, (iv)→S, (v)→T. (1 each = 5)

$(a+b)^2=a^2+2ab+b^2$
$(a-b)^2=a^2-2ab+b^2$
Sum of cubes and difference of cubes identities
Difference of squares

Section C

(Correct T/F = 1, valid reason = 1)

Q12. True. $2^2-4=4-4=0$ , so $x=2$ satisfies the equation. (2)

Q13. False. $5ab=5ba$ (multiplication is commutative), so they are like terms. (2)

Q14. True. $x^2+5x+6=(x+2)(x+3)$ , so $(x+3)$ is a factor. (2)

Q15. False. Dividing by $-2$ reverses the inequality: $x < -3$ . (2)

Q16. False. $\sqrt{16}+\sqrt9=4+3=7$ , but $\sqrt{25}=5$ ; $7\ne5$ . (2)

Q17. True. By Remainder Theorem remainder $=p(1)=1+1+1=3$ . (2)

Q18. True. Absolute value is never negative, so $|x|=-4$ has no solution. (2)

[
  {"claim":"Sum of roots of x^2-5x+6 is 5","code":"x=symbols('x'); r=solve(x**2-5*x+6,x); result=(sum(r)==5)"},
  {"claim":"(x+3) divides x^2+5x+6","code":"x=symbols('x'); result=(rem(x**2+5*x+6,x+3,x)==0)"},
  {"claim":"Remainder of x^2+x+1 by (x-1) is 3","code":"x=symbols('x'); p=x**2+x+1; result=(p.subs(x,1)==3)"},
  {"claim":"1/sqrt(3) rationalized equals sqrt(3)/3","code":"result=simplify(1/sqrt(3)-sqrt(3)/3)==0"}
]