Math Olympiad Test: Polynomials - 4 - Class 10 MCQ

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

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)

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

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

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

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

Now, 

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)

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

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

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