Class 10 Remainder and Factor Theorems Mock Test (ICSE)

Topics in Remainder and Factor Theorems

• Remainder theorem
• Factor theorem
• Factorising cubic polynomials
• Finding unknown coefficients

Remainder and Factor Theorems practice questions with answers

1. Q1Multiple choice·1 mark
   For $p(x) = x^3 - 2x^2 - 5x + 6$, substitution shows that $p(-2) = 0$, $p(1) = 0$ and $p(3) = 0$. Which of the following is a factor of $p(x)$?

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

   1. Correct option: (c) $(x - 3)$, since $p(3) = 0$ by the factor theorem+1

   Answer: (c)
2. Q2Multiple choice·1 mark
   When the polynomial $3x^3 - 5x^2 + 7x - 4$ is divided by $(x - 1)$, the remainder is:

   1. (a)$5$
   2. (b)$1$
   3. (c)$-1$
   4. (d)$3$
   Show answer and step marks

   1. Correct option: (b) $1$. By the Remainder Theorem, the remainder is $f(1) = 3(1)^3 - 5(1)^2 + 7(1) - 4 = 3 - 5 + 7 - 4 = 1$.+1

   Answer: (b)
3. Q3Short answer·2 marks
   Let $p(x) = 2x^3 + x^2 - 13x + 6$. (a) Explain why $(x-2)$ is a factor of $p(x)$, using the value of $p(2)$. [1] (b) Divide $p(x)$ by $(x-2)$ and write the quotient. [1]
   Show answer and step marks

   1. (a) $p(2) = 2(2)^3 + (2)^2 - 13(2) + 6 = 16 + 4 - 26 + 6 = 0$+0.5
   2. (a) Since the remainder on division by $(x-2)$ is zero, by the factor theorem $(x-2)$ is a factor of $p(x)$+0.5
   3. (b) Dividing: $2x^3 + x^2 - 13x + 6 = (x-2)(2x^2 + 5x - 3)$+1

   Answer: (a) $p(2) = 0$, so $(x-2)$ is a factor by the factor theorem. (b) Quotient $= 2x^2 + 5x - 3$
4. Q4Short answer·2 marks
   A rectangular wooden crate used to carry apples on a trip to a hill station has length, breadth and height, each a linear expression in $x$. Its volume, in cubic centimetres, is $V = x^3 + 9x^2 + 26x + 24$, where $x$ is a positive integer. Factorise $V$ completely.
   Show answer and step marks

   1. Putting $x = -2$: $(-2)^3 + 9(-2)^2 + 26(-2) + 24 = -8 + 36 - 52 + 24 = 0$, so $(x+2)$ is a factor+0.5
   2. Dividing $x^3 + 9x^2 + 26x + 24$ by $(x+2)$ gives the quotient $x^2 + 7x + 12$+0.5
   3. $x^2 + 7x + 12 = (x+3)(x+4)$, so $V = (x+2)(x+3)(x+4)$+1

   Answer: $V = (x+2)(x+3)(x+4)$ cm³
5. Q5Long answer·4 marks
   Let $f(x) = 2x^3 + px^2 - 3x + 4$, where $p$ is a constant. The remainder when $f(x)$ is divided by $(x - 2)$ is equal to the remainder when $f(x)$ is divided by $(x + 1)$. (a) Find the remainder when $f(x)$ is divided by $(x - 2)$, in terms of $p$. [1] (b) Find the remainder when $f(x)$ is divided by $(x + 1)$, in terms of $p$. [1] (c) Form an equation in $p$ using the given condition. [1] (d) Solve the equation to find the value of $p$. [1]
   Show answer and step marks

   1. (a) Remainder $= f(2) = 2(2)^3 + p(2)^2 - 3(2) + 4 = 14 + 4p$+1
   2. (b) Remainder $= f(-1) = 2(-1)^3 + p(-1)^2 - 3(-1) + 4 = -2 + p + 3 + 4 = p + 5$+1
   3. (c) Equating the two remainders: $14 + 4p = p + 5$+1
   4. (d) $3p = -9$, so $p = -3$+1

   Answer: (a) $14 + 4p$; (b) $p + 5$; (c) $14 + 4p = p + 5$; (d) $p = -3$
6. Q6Long answer·4 marks
   If $(x - 2)$ is a factor of the polynomial $f(x) = 2x^3 + ax^2 + bx - 6$ and when $f(x)$ is divided by $(x - 3)$, it leaves a remainder of $15$, find the values of $a$ and $b$.
   Show answer and step marks

   1. Since $(x - 2)$ is a factor of $f(x)$, $f(2) = 0$. Substituting $x = 2$ in $f(x)$: $2(2)^3 + a(2)^2 + b(2) - 6 = 0 \implies 16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5$.+1.5
   2. When $f(x)$ is divided by $(x - 3)$, the remainder is $15$, so $f(3) = 15$. Substituting $x = 3$: $2(3)^3 + a(3)^2 + b(3) - 6 = 15 \implies 54 + 9a + 3b - 6 = 15 \implies 9a + 3b = -33 \implies 3a + b = -11$.+1.5
   3. Subtract $2a + b = -5$ from $3a + b = -11$: $a = -6$.+0.5
   4. Substitute $a = -6$ in $2a + b = -5$: $-12 + b = -5 \implies b = 7$.+0.5

   Answer: $a = -6, b = 7$

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