Shown below is a part of the graph of a polynomial h(x).
On dividing h(x) by which of the following will the remainder be zero?
(i) (x – 2)
(ii) (x + 2)
(iii) (x – 4)
(iv) (x + 4)
Choose the correct option from the following:
only (ii)
only (i) and (iii)
only (ii) and (iv)
cannot be determined without knowing the polynomial h(x).
(1)
Since the given graph intersects the x-axis at x = –2, therefore (x + 2) is a factor of h(x).
So, on dividing h(x) by (x + 2), we get remainder zero
If the sum of zeros of the polynomial , then value of k is:
√2
2
2√2
1/2
(2)
Given polynomial is p(x) = 2x² – k√2x + 1.
Let α, β be its zeros then
α + β = –b/a = –(–k√2)/2 = k√2/2 = √2
[Here a = 2, b = –k√2, c = 1]
⇒ k√2 = 2√2 ⇒ k = 2
If α and β are the zeros of the polynomial p(x) = kx² – 30x + 45k and α + β = αβ, then the value of k is:
–2/3
–3/2
3/2
2/3
(4)
Given polynomial is p(x) = kx² – 30x + 45k.
We know α + β = –b/a and αβ = c/a
⇒ α + β = 30/k and αβ = 45k/k = 45
Given α + β = αβ ⇒ 30/k = 45 ⇒ k = 30/45 = 2/3
The zeros of the quadratic polynomial 2x² – 3x – 9 are:
3, –3/2
–3, –3/2
–3, 3/2
3, 3/2
(1)
Given quadratic polynomial is 2x² – 3x – 9.
For zeros, we have 2x² – 3x – 9 = 0 ⇒ 2x² – 6x + 3x – 9 = 0 ⇒ (x – 3)(2x + 3) = 0
⇒ x = 3 or x = –3/2
The number of quadratic polynomials having zeros –5 and –3 is:
1
2
3
more than 3
(4)
Given: zeros of the polynomial are –5 and –3.
Polynomials having zeros –5 and –3 are given by p(x) = k(x + 5)(x + 3).
⇒ p(x) = k(x² + 8x + 15), where k is any real constant.
∴ k can have infinitely many values. So, number of possible polynomials is infinitely many.
If one zero of the polynomial x² + 3x + k is 2, then the value of k is:
–10
10
5
–5
(1)
Given: one zero = 2.
∴ Substituting x = 2, we get 2² + 3(2) + k = 0 ⇒ 4 + 6 + k = 0 ⇒ k = –10
The degree of polynomial having zeros –3 and 4 only is:
2
1
more than 3
3
(1)
The degree of polynomial having zeros –3 and 4, i.e., only two zeros, is 2 because it will be a quadratic polynomial.
i.e., p(x) = (x – (–3))(x – 4) ⇒ p(x) = (x + 3)(x – 4) ⇒ p(x) = x² – x – 12, which is a polynomial of degree 2.
Which of these are the quotient and the remainder when (2x³ – 9x + 3x² + 12) is divided by (x – 1)?
quotient = (2x² – 7x – 4) and remainder = 8
quotient = (2x² + 7x + 4) and remainder = 16
quotient = (2x² + 5x – 4) and remainder = 8
quotient = (2x² + 5x + 4) and remainder = 16
(3)
Divide dividend by divisor, we get:
x – 1 | 2x³ + 3x² – 9x + 12
2x² + 5x – 4
So, the quotient is 2x² + 5x – 4 and remainder is 8.
Let p be a prime number. The quadratic equation having its roots as factors of p is:
x² – px + p = 0
x² – (p + 1)x + p = 0
x² + (p + 1)x + p = 0
x² – px + p + 1 = 0
(2)
Since p is a prime number, its factors are 1 and p itself. So, α = 1 and β = p.
General form of quadratic equation: x² – (sum of zeros)x + product of zeros = 0.
⇒ x² – (1 + p)x + p = 0 is the required quadratic equation.
If α and β are two zeros of a polynomial f(x) = px² – 2x + 3p and α + β = αβ, then p is:
–2/3
2/3
1/3
–1/3
(2)
Given polynomial is f(x) = px² – 2x + 3p.
So, α + β = –b/a = –(–2)/p = 2/p and αβ = c/a = 3p/p = 3 [a = p, b = –2, c = 3p].
Since α + β = αβ ⇒ 2/p = 3 ⇒ p = 2/3
