The zeroes of the polynomial are
(2)
Let
For zeros of
Its zeros are
If α and β are the zeros of a polynomial and α + β = αβ, then p is
(2)
Given,
Since and are the zeroes of the given polynomial.
(given)
The zeroes of a polynomial are twice the zeroes of the polynomial . The value of p is :
(1)
Given polynomials : ...(i)
and ...(ii)
Zero of polynomial are: and
Now, zero of polynomial are 4 and
Sum of zeroes
If the sum of zeroes of the polynomial is , then value of is :
(2)
Sum of zeroes
If and are zeroes of the polynomial , the value of is
−3/7
3/5
3/7
−5/7
(3)
Now,
Consider the polynomial Which of the following statements are true about its graph and zeros?
Choose the correct option from the following:
both (i) and (ii)
both (ii) and (iii)
(i), (ii) and (iv)
both (i) and (iv)
(3)
Factoring p(x) gives x(x – 1)(x – 2), indicating zeros at x = 0, 1, and 2. The graph intersects the x-axis at these points.
For the polynomial r(x) = x² – 4x + 4, which of the following statements are correct about its graph and zeros?

(i) The graph touches the x-axis at x = 2.
(ii) The graph intersects the x-axis at x = 2.
(iii) The polynomial is equivalent to (x – 2)².
(iv) The vertex of the graph is at the point (2, 0).
Choose the correct option from the following:
(i), (iii) and (iv)
(ii) and (iii)
(i), (ii) and (iv)
(i) and (iv)
(1)
Given polynomial r(x) can be rewritten as (x – 2)², indicating a repeated zero at x = 2. The graph touches the x-axis at this point and does not cross it. The vertex of the graph is at point (2, 0).
For the polynomial t(x) = x³ – 6x² + 11x – 6, which statements are true about its graph and zeros?
(i) The polynomial has a zero at x = 1.
(ii) The polynomial has a zero at x = 2.
(iii) The graph of the polynomial intersects the x-axis at x = 3.
(iv) The polynomial can be factorised as (x – 1)(x – 2)(x – 3).
Choose the correct option from the following:
(i), (iii) and (iv)
(i), (ii), (iii) and (iv)
(i), (ii) and (iv)
(i) and (iv)
(2)
At x = 1, t(1) = 1 – 6 + 11 – 6 = 0
So (x – 1) is a factor of t(x).
∴ t(x) = (x – 1)(x² – 5x + 6)
⇒ t(x) = (x – 1)(x – 2)(x – 3)
Factoring t(x) as (x – 1)(x – 2)(x – 3) reveals zeros at x = 1, 2, 3. The graph intersects the x-axis at these points.
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