RS Aggarwal Test: Polynomials - Class 10 MCQ

p(x) = x²- kx - 15

Given: p(5) = 0
⇒ (5)²- k(5) - 15 = 0
⇒ 25 - 5k - 15 = 0
⇒ 5k = 10
⇒ k = 10/5 = 2

Thus, Value of k is 2

Let one root be a, other root is a².
(x - a)(x - a²) = x² + (- a² - a)x + a³
p³ = (- a² - a)³ = - a⁶ - 3a⁵ - 3a⁴ - a³
- q(3p - 1) = - a³(3 (- a² - a) -1) = 3a⁵ + 3a⁴ + a³
q² = a⁶
Substituting the values
- a⁶ - 3a⁵ - 3a⁴ - a³ + 3a⁵ + 3a⁴ + a³ + a⁶ = 0

The correct answer is a.
From the polynomial we get,
α + β = - b/a = - (- 5/15) = 1/3
αβ = c/a = (6/15) = 2/5
To find: (1 + 1/α) (1 + 1/β)...... (taking L.C.M) 
= (α + 1/α) (β + 1/β) ....(simplifying brackets)
= αβ + α + β + 1 / αβ
=( 2/5 + 1/3+ 1)  /  2/5
=26/15 X 5/2 = 13/3

Let α and β be the roots of the given equation.
Then, α + β = 5/2 and αβ = 2/2 = 1
∴ 2α + 2β
∴ 2(α + β)
∴ 5,(2α)(2β) = 4
So, the requiared equation is : 
x² − 5x + 4 = 0

Let p(x) = ax + b
p(k) = ak + b = 0
∴ k is zero of p(x)

p(x) = x² + (a + 1)x + b
Given: zeros of polynomial is 2, -3
For x = 2
(2)² + (a + 1)² + b = 0
⇒ 4 + 2a + 2 + b = 0
⇒ 2a + b = -6 ...(i)

For x = -3
(-3)² + (a + 1) (-3) + b = 0
⇒ 9 - 3a - 3 + b = 0
⇒ -3a + b = -6 ..(ii)

Solving (i) and (ii), we get 5a = 0
⇒ a = 0 and b = -6

Alternative method:

p(x) = x² + (a + 1)x + b
Given: zeros of polynomial is 2, -3

Sum of zeros: -(a+1) = 2-3
⇒ a = 0
Product of zeros: b = 2.(-3) = -6 

 Let p (x) = ax² + bx + c be the required polynomial whose zeroes are -2 and 5.

∴ Sum of zeroes = -b/a
⇒  - 2 + 5 = 3 = -(-3)/1  ...(i)
and product of zeroes = c/a
⇒  -2 x 5 = -10/1 ...(ii)
From Eqs. (i) and (ii)
a = 1, b = -3 and c = -10
∴ p(x) = ax² + bx + c = 1.x² - 3x - 10
= x² - 3x - 10
But we know that, if we multiply/divide any polynomial by any arbitrary constant. Then, the zeroes of polynomial never change.
∴ p(x) = kx² - 3kx - 10k [where, k is a real number]
⇒ p(x) = x²/k - 3x/k -10/k [where, k is a non-zero real number]

Hence, the required number of polynomials are infinite i.e., more than 3.

For any quadratic polynomial ax² + bx + c, a≠0, graph for the corresponding equation:

• Has one of the two shapes either open upwards like ∪ or open downwards like  ∩ depending on whether a > 0 or a < 0. These curves are called parabolas. 
• The curve of a quadratic polynomial crosses the X-axis on at most two points.

Thus, option A is correct.

Let the roots be α and 1/α.

Product of roots = α(1/α) = 1
⇒ m/5 = 1
⇒ m = 5

Sum of zeroes = -b/a