Math Olympiad Test: Polynomials- 2 - Free MCQ with solutions Class 10

α, β are zeros of 2x² + 5x − 10
then ab = -10/2 = -5

Here l + m + n = p, lmn = r

Required polynomial
= x² − (α + β)x + αβ
⇒ x² − (−6)x + (−4) = x² + 6x − 4

(-x² + x - 1) - x³ + 3x² - 3x + 5(x - 2)

Given α and β are zeros of x² − px + d
then 
and 
∴ 

We have abx ² + (b²−ac)x − bc




The zeros of the given polynomial are  and 

Here α, β are the zeros of x² - 5x + P
then α + β = 5 …(1)
and αβ = P
α - β = 1 …(2)
Solving (1) and (2) we get
2α = 6 ⇒ α = 3
and β = 5 - 3 = 2
∴ αβ = P = (3) (2) = 6

Given 6x² + x − 2
If a, b are its zeros
then 
Now 

2 x⁴ − 3x³ − 3x² + 6x − 2
Let the other roots be α and β.
Sum of the roots = 
…(1)
Product of roots = -2/2 = -1 
√2 + (- √2)αβ  = -1 ⇒ αβ = 1/2…(2)
(α - β)² = (α + β)² - 4αβ

…(3)
From equation (1) and (3)

Let the zeros are α, -α, β.
∴ 
⇒ β = 5
and (α) (-α) (β) = 
⇒ -α²β = -80 ⇒ α²β = 80
⇒ 
⇒ α = ±4
The zeros are 4, -4, 5.

Given α + β + γ = 6
αβ + βγ + γα = -1
αβγ = -30
The cubic polynomial is
P(x) = x³ - (α + β + γ)x² + (αβ + βγ + γα)x - αβγ
= x³ - 6x² - x - (-30)
(substituting the given values)
= x³ - 6x² - x + 30

Polynomial = (−x² + x − 1) ( x − 2) + 3
= −x³ + 2x² + x² − 2x − x + 2 + 3
= −x³ + 3x² − 3x + 5

(3x² + 4 x + 1) 6 x⁴ + 8 x³ + 17x² + 21x + 7 (2x² + 5)


remainder = x + 2
ax + b = x + 2 (Given)
⇒ a = 1; b = 2

We have α + β + y = -2 - 3 - 1 = -6
αβ + βy + yα = (-2) (-3) + (-3) (-1) +(-1) (-2) = 6 + 3 + 2 = 11
and αβγ = (-2) (-3) (-1) = -6
∴ Required polynomial
= x³ - (-6)x² + 11x - (-6)
= x³ + 6x² + 11x + 6

Here, (2 − x + x²)(3x − 1)
= 6x − 2 − 3x² + x + 3x³ − x²
= 3x³ − 4x² + 7x − 2