Assertion & Reason Test: Polynomials - Grade 10 MCQ

Assertion (A): If the sum of the zeroes of the quadratic polynomial x² - 2kx + 8 is 2, then the value of
k is 1.
Reason (R): The sum of the zeroes of a quadratic polynomial ax² + bx + c is - (b/a).
Let's evaluate these statements:
For a quadratic equation ax2 + bx + c, the sum of the zeroes is given by - (b/a).
Given the quadratic polynomial x² - 2kx + 8, if the sum of the zeroes is 2, then we have:
- (- 2k / 1) = 2
2k = 2
k = 1
So, both assertion (A) and reason (R) are true, and reason (R) correctly explains assertion (A).
Therefore, the correct choice is:
1. Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Zeroes of f(x) = x² − 4x − 5 are obtained by solving:
x² − 4x − 5 = 0, which implies 
x² − 5x + x − 5 = 0 or, x(x − 5) + 1(x − 5) = 0 
which means x = 5 or x = −1 
Thus the assertion is correct. 
However, the reason is incorrect. 
The numbers given are not the zeroes  of x² − 4x + 7

Assertion (A): A polynomial of degree n can have more than n terms. For example, a
polynomial like xⁿ + x^n-1 + ... + x + 1 has n + 1 terms, which is more than n. So, the
assertion is false.
Reason (R): The number of coefficients in a polynomial is indeed one more than its degree.
This is because a polynomial of degree n can have terms ranging from x⁰ (constant term) to
xⁿ, and there are n + 1 coefficients, one for each term. This reason is true.

Let α, 1/α be the zeroes of p(x) then we have
Product of Zeroes
= a x (1/a) = 4k / (k² + 4) = 1
⇒ k² – 4k + 4 = 0
⇒ (k – 2)² = 0 ⇒ k = 2

Assertion: The polynomial x² - 5x + 6 can be factored as (x - 2)(x - 3).
Verification:
Let's expand (x - 2)(x - 3) to see if it equals x² - 5x + 6.
(x - 2)(x - 3) = x . x + x . (-3) - 2 . x - 2 . (-3)
= x² - 3x - 2x + 6
= x² - 5x + 6
Reason: The roots of the polynomial are 2 and 3, which can be used to express the polynomial in
its factored form.
Verification:
To find the roots of the polynomial x² - 5x + 6, we can set it equal to zero and solve for x:
x² - 5x + 6 = 0
Factoring the Quadratic:
We look for two numbers that multiply to 6 (constant term) and add up to -5 (coefficient of x).
The numbers are -2 and -3.
x2 - 2x - 3x + 6 = 0
x(x - 2) - 3(x - 2) = 0
(x - 2)(x - 3) = 0
Solutions:
x - 2 = 0 ⇒ x = 2
x - 3 = 0 ⇒ x = 3
Conclusion: The roots of the polynomial are 2 and 3. These roots are used to express the
polynomial in its factored form as (x - 2)(x - 3). Therefore, Reason (R) is correct.

Both the assertion and the reason are true, but the reason provided is related to the sum of the zeroes, not their product. The correct explanation involves the formula for the product of the zeroes of a quadratic polynomial, c/a, which would lead to 5k/−10, hence determining the value of k.

Answer: d) Assertion is incorrect, but Reason is correct.

Explanation:
A polynomial of degree 3 can have at most three real roots. The assertion stating that it can have at most two real roots is incorrect.

Option A: Both A and R are true and R is the correct explanation for A.
The polynomial p(x) = (x - 1)(x - 2)(x - 3) is factored into three linear factors: (x - 1) , (x - 2) , and  (x - 3) .
The zeroes of a polynomial are the values of  x  for which p(x) = 0. Setting each factor equal to zero gives the zeroes as x = 1 , x = 2 , and x = 3 , which confirms Assertion (A) as correct.
The reason (R) correctly explains the concept that the zeroes of a polynomial are the x-coordinates where the polynomial graph intersects or touches the x-axis, or where p(x) = 0.
Therefore, both the assertion and reason are correct, and the reason provides the correct explanation for the assertion.

- Assertion (A) is incorrect: The degree of a polynomial is not always 0. The degree of a polynomial is the highest power of the variable with a non-zero coefficient. 
- Reason (R) is incorrect: A polynomial can have exponents. These exponents are what define the degree of the polynomial.
- Correct Answer: Neither Assertion nor Reason is correct.