Q. 64.2( 11 Votes )
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.
Answer :
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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