Q. 64.2( 11 Votes )

# The two polynomials x^{3} + 2x^{2} – px – 7 and x^{3} + px^{2} – 12x + 6 are divided by (x + 1) and (x – 2) respectively and if the remainders R_{1} and R_{2} are obtained and if 2R_{1} + R_{2} = 6, then let us calculate the value of p.

Answer :

Let the polynomials be:

A(x) = x^{3} + 2x^{2} – px – 7

B(x) = x^{3} + px^{2} – 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 R_{1}.

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

To find zero,

⇒ x + 1 = 0

⇒ x = -1

The required remainder, R_{1} = A(-1).

⇒ R_{1} = (-1)^{3} + 2(-1)^{2} – p(-1) – 7

⇒ R_{1} = -1 + 2 + p – 7

⇒ R_{1} = 1 – 7 + p

⇒ R_{1} = p – 6 …(i)

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

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

To find zero,

⇒ x – 2 = 0

⇒ x = 2

The required remainder, R_{2} = B(2)

⇒ R_{2} = (2)^{3} + p(2)^{2} – 12(2) + 6

⇒ R_{2} = 8 + 4p – 24 + 6

⇒ R_{2} = 4p + 14 – 24

⇒ R_{2} = 4p – 10 …(ii)

We have, 2R_{1} + R_{2} = 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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