RD Sharma Test: Polynomials - Class 10 MCQ

If the graph of a polynomial intersects the x-axis at three points, then the number of zeroes are 3 because number of zeroes of the polynomial are the number of the coordinates of the points where its graph intersects the x-axis.

α + β = - b/a = - 5/1 And α​​​​​​​β = c/a = - 5/1
∴​​​​​​​ α​​​​​​​ + β​​​​​​​ = α​​​​​​​β

Sum of the zeroes of the polynomial = -b/a = 6/1 = 6 And Product of the zeroes of the polynomial = c/a = 8/1 = 8

A polynomial of degree 3 is called a cubic polynomial. 

For example, x³−1, 4a³ − 100a² + a − 6, and m²n + mn² are all cubic polynomials with atmost 3 zeroes (having degree 3).

Hence, the polynomial having atmost 3 zero is a cubic polynomial.

Here  α + β = - b/a = [-(- 5) / 1] α + β = 5 ……….(i)
And it is given that α−β = 1 ……….(ii)
On solving eq. (i) and eq. (ii),
we get α = 3, β​​​​​​​ = 2 ∴​​​​​​​ α​​​​​​​β = c/a
⇒ 3 x 2 = b/1
⇒ b = 6

Let α, β, γ are the zeroes of the given polynomial. Given: α = 2
Since αβγ = - d/a 
⇒ 2 x βγ = - 6/1
⇒ βγ = - 6/2 = -3

According to question, p(2) = 0 
⇒ (2)² + 3 × 2 + k = 0
⇒ 4 + 6 + k = 0
⇒ k = −10

Let α, β, γ are the zeroes of the given polynomial.
Since α + β + αβ = (−b/a) + (c/a) = (−b + c)/a
∴ α + β + αβ = (-5 + 1)/2 = - 4/2 = - 2

Let one zero of the given polynomial be athen the other zero be 1/α.
∵ αβ = c/α ∴ α x 1/α = 3k / k + 4
⇒ k + 4 = 3k ⇒ 2k = 4 ⇒ k = 2

 Given: α = 0, β = 0 and γ = ?
Since, α + β + γ = - b/a
∴ 0 + 0 + γ = - b/a
⇒ γ = - b/a

Let one zeroes of the given polynomial be α and β. According to the question, 

Sum of the zeroes = -b/a = 0

 ⇒ -(p-3)/2 = 0
 ⇒ - (p-3) = 0×2

 ⇒ - (p-3) = 0

 ⇒ p = - 3

Here α + β = -3 + 6 = 3/1 = -(- 3)/1 = b/a and αβ = (- 3) x 6 = -18/1 = c / a
on comparing we get, a = 1, b = -3, c = -18 Putting these values in the general form of quadratic polynomial
ax² + bx + c = x² - 3x - 18 = x²/6 - x/2 - 3 [Dividing all terms by 6] r

Given:  α + β​​​​​​​ = -b/a = √2 / 1 = -(- √2) / 1
= -(- 3 √2) / 3 And αβ = c/a = 1/3
On comparing, we get, a = 3, b = −3√2–, c = 1
Putting these values in general form of a quadratic polynomial ax² + bx + c, we have 3x² − 3√2x + 1

Let α, β are the zeroes of the given polynomial.
Given:  α + β = αβ ⇒ -b/a = c/a ⇒ -b = c
 ⇒ −(−27) = 3k² ⇒ k² = 0 ⇒ k= ±3

The term quadratic describes something that pertains to squares, to the operation of squaring, to terms of the second degree, or equations or formulas that involve such terms.It involves the second and no higher power of an unknown quantity or variable.