The two polynomials x³ + 2x² – px – 7 and x³ + px² – 12x + 6 are divided by (x + 1) and (x – 2) respectively and if the remainders R₁ and R₂ are obtained and if 2R₁ + R₂ = 6, then let us calculate the value of p.

Let the polynomials be:

A(x) = x³ + 2x² – px – 7

B(x) = x³ + px² – 12x + 6

We will use remainder theorem here.

By Remainder theorem that says, f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

When A(x) is divided by (x + 1), it will leave remainder R₁.

Let’s also find zero of linear polynomial, (x + 1).

To find zero,

⇒ x + 1 = 0

⇒ x = -1

The required remainder, R₁ = A(-1).

⇒ R₁ = (-1)³ + 2(-1)² – p(-1) – 7

⇒ R₁ = -1 + 2 + p – 7

⇒ R₁ = 1 – 7 + p

⇒ R₁ = p – 6 …(i)

When B(x) is divided by (x – 2), it will leave remainder R₂.

Let’s also find zero of linear polynomial, (x – 2).

To find zero,

⇒ x – 2 = 0

⇒ x = 2

The required remainder, R₂ = B(2)

⇒ R₂ = (2)³ + p(2)² – 12(2) + 6

⇒ R₂ = 8 + 4p – 24 + 6

⇒ R₂ = 4p + 14 – 24

⇒ R₂ = 4p – 10 …(ii)

We have, 2R₁ + R₂ = 6

Substituting values from (i) and (ii), we get

2(p – 6) + (4p – 10) = 6

⇒ 2p – 12 + 4p – 10 = 6

⇒ 6p – 22 = 6

⇒ 6p = 6 + 22

⇒ 6p = 28

⇒

⇒

Thus, p = 14/3.