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CBSE Class 10  >  Class 10 Test  >  Olympiad Preparation  >  Math Olympiad Test: Polynomials - 4 - Class 10 MCQ

Math Olympiad Test: Polynomials - 4 - Free MCQ with solutions Class 10


MCQ Practice Test & Solutions: Math Olympiad Test: Polynomials - 4 (10 Questions)

You can prepare effectively for Class 10 Olympiad Preparation for Class 10 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Math Olympiad Test: Polynomials - 4". These 10 questions have been designed by the experts with the latest curriculum of Class 10 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 10

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Math Olympiad Test: Polynomials - 4 - Question 1
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A polynomial of the form ax5 + bx3 + cx2 + dx + e has atmost _______ zeroes.

Detailed Solution: Question 1

Since, degree of given polynomial is 5,
so ax5 + bx3 + cx2 + dx + e has atmost 5 zeroes.

Math Olympiad Test: Polynomials - 4 - Question 2
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If one zero of the polynomial f (x) = (k2 + 4) x2 + 13x + 4k is reciprocal of the other, then k is equal to _______.

Detailed Solution: Question 2

f(x) = (k2 + 4)x2 + 13x + 4k
Now, let a and b be the roots, then according to the question, α = 1/β ⇒ αβ = 1
Now, we know that αβ = 4k/(k2 + 4)

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Math Olympiad Test: Polynomials - 4 - Question 3
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For x2 + 2x + 5 to be a factor of x4 + αx2 + β, the values of α and β should respectively be _______.

Detailed Solution: Question 3

For x2 + 2x + 5 to be a factor of x4 + αx2 + β, remainder should be zero.

Now, remainder should be equal to zero.
∴ α – 6 = 0 and β – 25 = 0
⇒ α = 6 and β = 25

Math Olympiad Test: Polynomials - 4 - Question 4
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If α and β are the roots of the equation 2x2 – 7x + 8 = 0, then the equation whose roots are(3α – 4β) and (3β – 4α) is _______.

Detailed Solution: Question 4

2x2 – 7x + 8 = 0
α, β are roots of above equation (ax2 + bx + c = 0)
∴ α + β = −b/a​ = 7/2 -(1)
αβ = 8/2​(c/a​) = 4

Roots of other eqn ⇒ (3α − 4β), (3β − 4α)

⇒ sum of roots = 3α − 4β + 3β − 4α
= −α − β = −(α + β)
=(−7/2​) (from (1))
⇒ product of roots ⇒ (3α − 4β)(3β − 4α)
= 9αβ − 12α2 − 12β2 + 16αβ
= 25αβ − 12(α2 + β2)
= 25αβ − 12((α + β)2 − 2αβ)
= 25αβ − 12(α + β)2 + 21αβ
= 49αβ − 12(α + β)2
⇒ 49(4) − 12(7​/2)2
⇒ 49(4) − 12 × 49/4 ​= 49(4 − 3)
⇒ (49)
⇒ quadratic poly ⇒ x2 - (sum of roots)x +(product of roots)
⇒ x2 − (2 − 7​)x + 49
⇒ 2x2 + 7x + 98

Math Olympiad Test: Polynomials - 4 - Question 5
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Area of a triangular field is (x4 – 6x3 – 26x2 + 138x – 35) m2 and base of the triangular field is (x2 – 4x + 1) m. Find the height of the triangular field.

Detailed Solution: Question 5

Base of the triangular field = (x2 – 4x + 1) m
Area of the triangular field = (1/2) x Base x Height
Now, x4 – 6x3 – 26x2 + 138x – 35 = 1/2 x (x2 − 4x + 1) × Height
⇒ Height = 

∴ Height = 2(x2 – 2x – 35) m

Math Olympiad Test: Polynomials - 4 - Question 6
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If α, β be two zeroes of the quadratic polynomial ax2 + bx – c = 0, then find the value of 

Detailed Solution: Question 6

Since, a and b are the zeroes of quadratic equation ax2 + bx – c = 0

Now, 

Math Olympiad Test: Polynomials - 4 - Question 7
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Raghav had ₹(6x3 + 2x2 + 3x) and he bought (4x2 + 3) shirts. The price of each shirt is ₹(x + 5). How much money is left with Raghav?

Detailed Solution: Question 7

Total amount Raghav had = ₹(6x3 + 2x2 + 3x)
Cost of one shirt = ₹(x + 5)
Number of shirts he bought = 4x2 + 3
∴ Amount spent by him = ₹(x + 5)(4x2 + 3)
= ₹(4x3 + 20x2 + 3x + 15)
Hence, money left with Raghav
= ₹(6x3 + 2x2 + 3x – 4x3 – 20x2 – 3x – 15)
= ₹(2x3 – 18x2 – 15)

Math Olympiad Test: Polynomials - 4 - Question 8
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A rectangular garden of length (2x3 + 5x2 – 7) m has the perimeter (4x3 – 2x2 + 4) m. Find the breadth of the garden.

Detailed Solution: Question 8

Length of the garden = (2x3 + 5x2 – 7) m
Perimeter of the garden = 2 × (length + breadth)
∴ 4x3 - 2x2 + 4 = 2(2x3 + 5x2 – 7 + breadth)
⇒ 2x3 – x2 + 2 = (2x3 + 5x2 – 7) + breadth
So, breadth of the rectangle
= 2x3 – x2 + 2 – 2x3 – 5x2 + 7 = (–6x2 + 9) m

Math Olympiad Test: Polynomials - 4 - Question 9
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Length and breadth of a rectangular park are (3x2 + 2x) m and (2x3 – 3) m respectively. Find the area of the park, when x = 3.

Detailed Solution: Question 9

Length of rectangular park = (3x2 + 2x) m
Breadth of rectangular park = (2x3 – 3) m
Area of park = length × breadth
= (3x2 + 2x) (2x3 – 3) = (6x5 + 4x4 – 9x2 – 6x) m
For x = 3, 6x5 + 4x4 – 9x2 – 6x
= 6 × 243 + 4 × 81 – 9 × 9 – 6 × 3 = 1683
Hence, area of park = 1683 m2

Math Olympiad Test: Polynomials - 4 - Question 10
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Two different container contains (2x3 + 2x2 + 3x + 3) L and (4x3 – 2x2 + 6x – 3) L water. What is biggest measure that can measure both quantities exactly?

Detailed Solution: Question 10

Capacity of both the containers is (2x3 + 2x2 + 3x + 3) L and (4x3 – 2x2 + 6x – 3) L
i.e., (2x2 + 3)(x + 1) L and (2x2 + 3)(2x – 1) L
Required measure is the H.C.F. of capacity of both the containers i.e., (2x2 + 3) L

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About this Class 10 Practice Test

This test contains 10 MCQs on Math Olympiad Test: Polynomials - 4 with detailed solutions for each question. It is part of the Olympiad Preparation for Class 10 course on EduRev, designed to align with the current Class 10 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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