Remainder & Factor Theorem — Assertion-Reason Type Questions

Assertion (A): When a polynomial f(x) is divided by (3x + 4), then the remainder is $f\Big(-\dfrac{3}{4}\Big)$.

Reason (R): Remainder theorem states that when a polynomial f(x) is divided by (x - α), then the remainder is f(α).

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

Remainder Theorem states that when a polynomial f(x) is divided by a linear factor (x − α), the remainder is f(α).

∴ Reason (R) is true.

Divisor :

⇒ 3x + 4 = 0

⇒ 3x = -4

⇒ x = $-\dfrac{4}{3}$

Thus, when a polynomial f(x) is divided by (3x + 4), then the remainder is $f\Big(-\dfrac{4}{3}\Big)$.

∴ Assertion (A) is false.

Hence, option 4 is the correct option.

Assertion (A): (x - 1) is a factor of x³ + 2x² - x - 2.

Reason (R): If (x + α) is a factor of f(x), then f(α) = 0.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

Let,

⇒ f(x) = x³ + 2x² - x - 2

⇒ f(1) = (1)³ + 2(1)² - 1 - 2

= 1 + 2 - 1 - 2

= 3 - 3

= 0.

Since, f(1) = 0.

Thus, (x − 1) is a factor of f(x) = x³ + 2x² - x - 2 if f(1) = 0.

∴ Assertion (A) is true.

⇒ x + a = 0

⇒ x = -a.

If (x + a) is a factor of f(x), then f(−a) = 0.

∴ Reason (R) is false.

A is true, R is false.

Hence, option 3 is the correct option.

Assertion (A): If (2x - 1) is a factor of polynomial f(x), then $f\Big(\dfrac{1}{2}\Big)$ = 0.

Reason (R): (ax + b) is a factor of f(x) implies $f\Big(-\dfrac{b}{a}\Big)$ = 0.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

By factor theorem,

If (x - a) is a factor of f(x), then f(a) = 0.

Given,

⇒ 2x - 1 = 0

⇒ 2x = 1

⇒ x = $\dfrac{1}{2}$

Thus, if (2x - 1) is a factor of polynomial f(x), then $f\Big(\dfrac{1}{2}\Big)$ = 0.

∴ Assertion (A) is true.

⇒ ax + b = 0

⇒ ax = -b

⇒ x = $-\dfrac{b}{a}$

Thus, if (ax + b) is a factor of polynomial f(x), then $f\Big(-\dfrac{b}{a}\Big)$ = 0.

∴ Reason (R) is true.

Thus, both A and R are true, and R is the correct explanation of A.

Hence, option 1 is the correct option.

Assertion (A): x³ + 2x² - x - 2 is a polynomial of degree 3.

Reason (R): x + 2 is a factor of the polynomial.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

Since, the highest value of power of x in the polynomial x³ + 2x² - x - 2 is 3.

Thus,

x³ + 2x² - x - 2 is a polynomial of degree 3.

∴ Assertion (A) is true.

By factor theorem,

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

x + 2 = 0

x = -2

Substituting x = -2 in x³ + 2x² - x - 2, we get :

⇒ (-2)³ + 2(-2)² - (-2) - 2

⇒ -8 + 2(4) + 2 - 2

⇒ -8 + 8 + 2 - 2

⇒ 0.

∴ Reason (R) is true.

Both A and R are true, but R is not the correct explanation of A.

Hence, Option 2 is the correct option.