Class 9 Factorisation Mock Test (ICSE)

Topics in Factorisation

• Taking out common factors
• Grouping
• Difference of two squares
• Splitting the middle term
• Sum and difference of two cubes

Factorisation practice questions with answers

1. Q1Multiple choice·1 mark
   The factorised form of $8x^{3} - 27y^{3}$ is:

   1. (a)$(2x-3y)(4x^{2}-6xy+9y^{2})$
   2. (b)$(2x+3y)(4x^{2}-6xy+9y^{2})$
   3. (c)$(2x-3y)(4x^{2}+6xy+9y^{2})$
   4. (d)$(2x-3y)(4x^{2}-9y^{2})$
   Show answer and step marks

   1. Correct option: (c) $(2x-3y)(4x^{2}+6xy+9y^{2})$, since $8x^{3} - 27y^{3} = (2x)^{3} - (3y)^{3}$+1

   Answer: (c)
2. Q2Multiple choice·1 mark
   The expression $ab + bc + ad + cd$ factorised by grouping is:

   1. (a)$(a + c)(b + d)$
   2. (b)$(a - c)(b - d)$
   3. (c)$(a + b)(c + d)$
   4. (d)$(a + d)(b + c)$
   Show answer and step marks

   1. Correct option: (a) Grouping terms as $b(a + c) + d(a + c)$ and taking out $(a + c)$ yields $(a + c)(b + d)$.+1

   Answer: (a)
3. Q3Short answer·2 marks
   Factorise completely: $2x^2 + 7x + 6$
   Show answer and step marks

   1. Split the middle term $7x$ into $4x$ and $3x$ since $2 \times 6 = 12$, $4 \times 3 = 12$ and $4 + 3 = 7$: $2x^2 + 4x + 3x + 6$+1
   2. Factor by grouping: $2x(x + 2) + 3(x + 2) = (2x + 3)(x + 2)$+1

   Answer: $(2x + 3)(x + 2)$
4. Q4Short answer·2 marks
   On the advice of her doctor, Meena walks every morning for exercise on a rectangular path in her home garden. The area of this path is $12x^2 + 18xy$ square metres, where $x$ and $y$ are in metres. (a) Factorise $12x^2 + 18xy$ by taking out the common factors. [1] (b) Find the area of the path, in square metres, when $x = 2$ m and $y = 1$ m, using your factorised form. [1]
   Show answer and step marks

   1. $12x^2 + 18xy = 6x(2x + 3y)$+1
   2. Substituting $x = 2$, $y = 1$: $6 \times 2 \times (2 \times 2 + 3 \times 1) = 12 \times 7 = 84$ m²+1

   Answer: (a) $6x(2x + 3y)$; (b) $84$ m²
5. Q5Long answer·4 marks
   Factorise $3x^2 + 13x + 14$ by splitting the middle term. (a) State two integers whose product is $3 \times 14$ and whose sum is $13$. [1] (b) Using the integers in (a), write the middle term $13x$ as the sum of two terms. [1] (c) Factorise $3x^2 + 13x + 14$ completely by grouping the four terms. [1] (d) Give a reason why the product of the two integers in (a) must equal $3 \times 14$. [1]
   Show answer and step marks

   1. (a) Two integers with product $42$ and sum $13$ are $6$ and $7$+1
   2. (b) $13x = 6x + 7x$+1
   3. (c) $3x^2 + 6x + 7x + 14 = 3x(x+2) + 7(x+2) = (x+2)(3x+7)$+1
   4. Reason: the split $px + qx$ must satisfy $pq = 3 \times 14$ so that the two groups $3x^2 + px$ and $qx + 14$ have the same bracket $(x + 2)$ and can be factorised together+1

   Answer: (a) $6$ and $7$; (b) $13x = 6x + 7x$; (c) $(x+2)(3x+7)$; (d) the product $pq$ must equal $ac = 3 \times 14 = 42$ so that the two groups share a common bracket
6. Q6Long answer·4 marks
   Consider the expression $2x^3 + 3x^2 - 8x - 12$. (a) Factorise the expression by grouping its four terms. [2] (b) Factorise the result of part (a) completely. [1] (c) Find the positive value of $x$ for which $2x^3 + 3x^2 - 8x - 12 = 0$. [1]
   Show answer and step marks

   1. Group the terms: $(2x^3 + 3x^2) + (-8x - 12)$+0.5
   2. Factor each group: $x^2(2x + 3) - 4(2x + 3)$+0.5
   3. Take out the common factor $(2x + 3)$: $(2x + 3)(x^2 - 4)$+1
   4. $x^2 - 4 = (x - 2)(x + 2)$, so the expression is $(2x + 3)(x - 2)(x + 2)$+1
   5. $(2x + 3)(x - 2)(x + 2) = 0$, so $x = -\dfrac{3}{2}$, $x = 2$ or $x = -2$+0.5
   6. The only positive value is $x = 2$+0.5

   Answer: (a) $(2x + 3)(x^2 - 4)$; (b) $(2x + 3)(x - 2)(x + 2)$; (c) $x = 2$

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