Factorisation — Assertion-Reason Type Questions

Given f(x) = ax² + bx + c, a ≠ 0, b, c ∈ R.

Assertion (A): The factorisation of ax² + bx + c is possible only if its discriminant = b² - 4ac < 0.

Reason (R): To factorise ax² + bx + c, split the coefficient of x into two real numbers such that their algebraic sum is b and their product is ac.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given f(x) = ax² + bx + c, a ≠ 0, b, c ∈ R.

Discriminant (D) = b² - 4ac

We know that,

For D ≥ 0, the equation has real roots so factorisation is possible.

So, assertion (A) is false.

To factorise ax² + bx + c, split the coefficient of x into two real numbers such that their algebric sum is b and their product is ac.

This describes the method of splitting the middle term, which is a valid technique used to factor quadratic expressions over the real numbers — provided real roots exist.

So, reason (R) is true.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Given a polynomial f(x) = 2x³ - 7x² - 5x + 4

Assertion (A): (x - 1) is not factor of f(x).

Reason (R): f(1) = -6.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given a polynomial f(x) = 2x³ - 7x² - 5x + 4

By the Factor Theorem, (x - r) is a factor of f(x) if and only if f(r) = 0.

(x - 1) is a factor of f(x) if and only if f(1) = 0.

f(1) = 2 . (1)³ - 7 . (1)² - 5 . (1) + 4

= 2 - 7 - 5 + 4 = -6.

Thus, (x - 1) is not a factor of f(x).

So, Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given f(x) = 16x³ - 8x² + 4x + 7

Assertion (A): When we subtract 1 from f(x), the resulting polynomial is divisible by (2x + 1).

Reason (R): $f\Big(-\dfrac{1}{2}\Big)$ = 1.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given f(x) = 16x³ - 8x² + 4x + 7

Subtract 1 from f(x), we get :

g(x) = 16x³ - 8x² + 4x + 6

By Factor Theorem,

(x - r) is a factor of f(x) if and only if f(r) = 0.

(2x + 1) is a factor of g(x) if and only if $g\Big(-\dfrac{1}{2}\Big)$ = 0.

$g\Big(-\dfrac{1}{2}\Big) = 16 \times \Big(-\dfrac{1}{2}\Big)^3 - 8 \times \Big(-\dfrac{1}{2}\Big)^2 + 4 \times \Big(-\dfrac{1}{2}\Big) + 6\\[1em] = 16 \times \Big(-\dfrac{1}{8}\Big) - 8 \times \Big(\dfrac{1}{4}\Big) + 4 \times \Big(-\dfrac{1}{2}\Big) + 6\\[1em] = -2 - 2 - 2 + 6\\[1em] = 0.$

So, assertion (A) is true.

$f\Big(-\dfrac{1}{2}\Big) = 16 \times \Big(-\dfrac{1}{2}\Big)^3 - 8 \times \Big(-\dfrac{1}{2}\Big)^2 + 4 \times \Big(-\dfrac{1}{2}\Big) + 7\\[1em] = 16 \times \Big(-\dfrac{1}{8}\Big) - 8 \times \Big(\dfrac{1}{4}\Big) + 4 \times \Big(-\dfrac{1}{2}\Big) + 7\\[1em] = -2 - 2 - 2 + 7\\[1em] = 1$

So, reason (R) is true.

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.