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Class 9 Exam  >  Class 9 Tests  >  Mathematics (Maths) Class 9  >  Test: Factorisation of Polynomials - Class 9 MCQ

Test: Factorisation of Polynomials - Class 9 MCQ


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10 Questions MCQ Test Mathematics (Maths) Class 9 - Test: Factorisation of Polynomials

Test: Factorisation of Polynomials for Class 9 2025 is part of Mathematics (Maths) Class 9 preparation. The Test: Factorisation of Polynomials questions and answers have been prepared according to the Class 9 exam syllabus.The Test: Factorisation of Polynomials MCQs are made for Class 9 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Factorisation of Polynomials below.
Solutions of Test: Factorisation of Polynomials questions in English are available as part of our Mathematics (Maths) Class 9 for Class 9 & Test: Factorisation of Polynomials solutions in Hindi for Mathematics (Maths) Class 9 course. Download more important topics, notes, lectures and mock test series for Class 9 Exam by signing up for free. Attempt Test: Factorisation of Polynomials | 10 questions in 10 minutes | Mock test for Class 9 preparation | Free important questions MCQ to study Mathematics (Maths) Class 9 for Class 9 Exam | Download free PDF with solutions
Test: Factorisation of Polynomials - Question 1
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Which of the following are the factors of a2 + ab +bc + ca

Detailed Solution for Test: Factorisation of Polynomials - Question 1

Factors of a2 + ab + bc + ca
• Step 1: Factor out common terms
Given expression: a2 + ab + bc + ca
= a(a + b) + c(b + a)
= a(a + b) + c(a + b)
• Step 2: Factor by grouping
= (a + c)(a + b)
Therefore, the factors of a2 + ab + bc + ca are (a + c)(a + b).

Test: Factorisation of Polynomials - Question 2
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Factoring 3x2-5x+2

Detailed Solution for Test: Factorisation of Polynomials - Question 2

3x² - 5x + 2
= 3x² - 2x - 3x + 2
= (3x² - 2x) + (-3x + 2)
= x(3x - 2) - 1(3x - 2)
= x(3x - 2) - 1(3x - 2)
= (x - 1)(3x - 2)

Therefore, the correct answer is A.

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Test: Factorisation of Polynomials - Question 3
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The number of zeros of x2 + 4x + 2      

Detailed Solution for Test: Factorisation of Polynomials - Question 3

To find the number of zeros of the quadratic equation x2 - 4x + 2, follow these steps:

  • The equation is in the form of ax2 + bx + c, where:
    a = 1
    b = -4
    c = 2
  • Use the quadratic formula: x = (-b ± √(b2 - 4ac)) / 2a.
  • Calculate the discriminant (b2 - 4ac):
    (-4)2 - 4(1)(2) = 16 - 8 = 8
  • Since the discriminant is positive, there are two real zeros.

Therefore, Option B is the correct answer.

Test: Factorisation of Polynomials - Question 4
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Factorise by splitting the middle term 2x2 - 11x + 12
Detailed Solution for Test: Factorisation of Polynomials - Question 4

(2x-3)(x-4)

= 2x² - 11x + 12

= 2x² - 8x - 3x + 12

= 2x(x-4) - 3(x-4)

= (2x-3)(x-4)

Test: Factorisation of Polynomials - Question 5
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The volume of cuboid is given by the expression x3+2x2-x-2. The dimension of the cuboid for x = 5 is:​
Detailed Solution for Test: Factorisation of Polynomials - Question 5

x³ + 2x² - x - 2
= x²(x + 2) - 1(x + 2)
= (x + 2)(x² - 1)
= (x + 2)(x - 1)(x + 1)

For x = 5, dimensions will be (5 + 2), (5 - 1), (5 + 1)

So, the dimensions of the cuboid are 7, 4, 6.

Test: Factorisation of Polynomials - Question 6
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Expansion of 2x(x + 2y) + 3x(2x - 3y) yields
Detailed Solution for Test: Factorisation of Polynomials - Question 6

The given expression is 2x(x + 2y) + 3x(2x - 3y).
To expand this expression, we distribute the terms inside the brackets:

  • 2x(x + 2y) = 2x² + 4xy
  • 3x(2x - 3y) = 6x² - 9xy

Adding the two results together we get:
2x² + 4xy + 6x² - 9xy = 8x² - 5xy
Therefore, the correct expansion of the expression is 8x² - 5xy, which matches option A.

Test: Factorisation of Polynomials - Question 7
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What are the two factors of quadratic polynomial x2-16x+64?
Detailed Solution for Test: Factorisation of Polynomials - Question 7

x² - 16x + 64
= x² - 8x - 8x + 64
= x(x - 8) - 8(x - 8)
= (x - 8)(x - 8)
= (x - 8)²

Test: Factorisation of Polynomials - Question 8
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Factorise the quadratic polynomial by splitting the middle term:
x2 + 14x + 45
Detailed Solution for Test: Factorisation of Polynomials - Question 8

To factorise the quadratic polynomial x2 + 14x + 45, follow these steps:

  • Identify two numbers that multiply to 45 and add up to 14.
  • The numbers are 9 and 5.
  • Rewrite the middle term: x2 + 9x + 5x + 45.
  • Factor by grouping:
    Group the first two and last two terms: (x2 + 9x) + (5x + 45).
    Factor out the common terms: x(x + 9) + 5(x + 9).
    Combine to get: (x + 9)(x + 5).

Thus, the factorised form is (x + 9)(x + 5).

Test: Factorisation of Polynomials - Question 9
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Factorise the quadratic polynomial by splitting the middle term: y2 – 4 y –21​
Detailed Solution for Test: Factorisation of Polynomials - Question 9

y2 – 4 y –21​
= y2 – 7 y + 3 y –21​
= y (y – 7) – 3 (y – 7)
= (y –​ 7) (y – 3)

Test: Factorisation of Polynomials - Question 10
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What is the value of p if x-2 is a factor of x2 – 6x + p ?
Detailed Solution for Test: Factorisation of Polynomials - Question 10

P(x) = ×2 - 6×+p
p (2)= (2)2-6 (2) +p
4- 12 + p = 0
-8 + p = 0
p = 8 

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