# 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.

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