Class 9 Expansions Mock Test (ICSE)

Topics in Expansions

• Square of a binomial
• Square of a trinomial
• Cube of a binomial
• Product of sum and difference
• Using expansions to evaluate expressions

Expansions practice questions with answers

1. Q1Multiple choice·1 mark
   The expansion of $(2x - 1)^{3}$ is:

   1. (a)$8x^{3} - 12x^{2} + 6x - 1$
   2. (b)$8x^{3} - 12x^{2} + 6x + 1$
   3. (c)$8x^{3} - 4x^{2} + 2x - 1$
   4. (d)$8x^{3} + 12x^{2} + 6x + 1$
   Show answer and step marks

   1. Correct option: (a) $8x^{3} - 12x^{2} + 6x - 1$+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   In the expansion of $(2a + 3)^3$, the coefficient of $a^2$ is:

   1. (a)$36$
   2. (b)$12$
   3. (c)$54$
   4. (d)$24$
   Show answer and step marks

   1. Correct option: (a) $36$+1

   Answer: (a)
3. Q3Short answer·2 marks
   (a) Expand $(4p - 7)(4p + 7)$. [1] (b) Give a reason why the expansion of $(4p - 7)(4p + 7)$ has no middle term, whereas $(4p - 7)^2$ has one. [1]
   Show answer and step marks

   1. Using $(a + b)(a - b) = a^2 - b^2$ with $a = 4p$ and $b = 7$: $(4p)^2 - 7^2$+0.5
   2. Simplify: $16p^2 - 49$+0.5
   3. The two middle terms are $+28p$ and $-28p$, which have opposite signs and cancel each other, so no middle term remains; $(4p - 7)^2$ has the middle term $-56p$ because both factors have the same sign pattern+1

   Answer: (a) $16p^2 - 49$; (b) the middle terms $+28p$ and $-28p$ cancel, since the factors differ only in the sign of the second term
4. Q4Short answer·2 marks
   Let $a$, $b$ and $c$ be numbers such that $a + b + c = 6$ and $a^2 + b^2 + c^2 = 20$. (a) Write the expansion of $(a + b + c)^2$. [1] (b) Using the expansion in part (a), find the value of $ab + bc + ca$. [1]
   Show answer and step marks

   1. $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$+1
   2. Substituting: $36 = 20 + 2(ab + bc + ca)$, so $2(ab + bc + ca) = 16$ and $ab + bc + ca = 8$+1

   Answer: (a) $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$; (b) $ab + bc + ca = 8$
5. Q5Long answer·4 marks
   At a science exhibition stall, a square chart panel is displayed on a stand. The side length of the panel, in centimetres, is $(3x - 4)$, where $x$ is a number chosen by the organisers. (a) Write the area of the panel in expanded form, in terms of $x$, in cm². [2] (b) Find the area of the panel, in cm², when $x = 5$. [1] (c) Find the value of $x$ for which the area of the panel is $25$ cm², given that the side length of the panel is a positive number. [1]
   Show answer and step marks

   1. Area = (side)² = $(3x - 4)^2$+0.5
   2. $= (3x)^2 - 2(3x)(4) + 4^2$+0.5
   3. $= 9x^2 - 24x + 16$+1
   4. Substitute $x = 5$: $9(25) - 24(5) + 16 = 225 - 120 + 16$+0.5
   5. Area $= 121$ cm²+0.5
   6. Side $= \sqrt{25} = 5$ cm, as the side is positive; so $3x - 4 = 5$+0.5
   7. $3x = 9$, so $x = 3$+0.5

   Answer: (a) $9x^2 - 24x + 16$ cm²; (b) $121$ cm²; (c) $x = 3$
6. Q6Long answer·4 marks
   Let $p$ and $q$ be two positive numbers such that $p + q = 8$ and $pq = 12$. (a) Write the expansion of $(p + q)^2$ in terms of $p$ and $q$. [1] (b) Using the expansion in (a), find the value of $p^2 + q^2$. [1] (c) Find the value of $(p - q)^2$. [1] (d) Given that $p > q$, find the value of $p - q$. [1]
   Show answer and step marks

   1. (a) $(p+q)^2 = p^2 + 2pq + q^2$+1
   2. (b) Substituting $p + q = 8$ and $pq = 12$: $64 = p^2 + 24 + q^2$, so $p^2 + q^2 = 64 - 24 = 40$+1
   3. (c) $(p-q)^2 = p^2 + q^2 - 2pq = 40 - 24 = 16$+1
   4. (d) $(p-q)^2 = 16$ and $p > q$ give $p - q = 4$+1

   Answer: (a) $(p+q)^2 = p^2 + 2pq + q^2$; (b) $p^2 + q^2 = 40$; (c) $(p-q)^2 = 16$; (d) $p - q = 4$

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