Let f(x) be a polynomial for which the remainders when divided by x – 1, x – 2, x – 3 respectively 3, 7, 13. Then the remainder of f(x) when divided by (x – 1) (x – 2) (x – 3) isa)f(x)b)x2 + x +1c)x2 +1d)x + 2Correct answer is option 'B'. Can you explain this answer?

Question

Let f(x) be a polynomial for which the remainders when divided by x &#8211; 1, x &#8211; 2, x &#8211; 3 respectively 3, 7, 13. Then the remainder of f(x) when divided by (x &#8211; 1) (x &#8211; 2) (x &#8211; 3) isa)f(x)b)x2 + x +1c)x2 +1d)x + 2Correct answer is option 'B'. Can you explain this answer?

Answer

Solution:

Given f(x) is a polynomial for which the remainders when divided by x - 1, x - 2, x - 3 respectively 3, 7, 13.

Let's assume the polynomial f(x) can be written as:

f(x) = q1(x)(x - 1) + 3
f(x) = q2(x)(x - 2) + 7
f(x) = q3(x)(x - 3) + 13

where q1(x), q2(x), q3(x) are quotients of the polynomial.

Expanding (x - 1)(x - 2)(x - 3):

(x - 1)(x - 2)(x - 3) = (x^2 - 3x + 2)(x - 3)
= x^3 - 6x^2 + 11x - 6

Using Remainder theorem:

The remainder when f(x) is divided by (x - 1)(x - 2)(x - 3) is given by:

f(x) = (x - 1)(x - 2)(x - 3)q(x) + r(x)

where q(x) is the quotient and r(x) is the remainder.

Now, we need to find the value of r(x).

Substituting the given equations:

Let's substitute the given equations of f(x) in the above equation,

f(x) = (x - 1)(x - 2)(x - 3)q(x) + r(x)
f(x) = (x - 2)(x - 3)q1(x) + 3
f(x) = (x - 1)(x - 3)q2(x) + 7
f(x) = (x - 1)(x - 2)q3(x) + 13

Finding r(x):

Now, we need to substitute the values of x = 1, x = 2, and x = 3 in the above equations one by one to find the value of r(x).

Substituting x = 1,

r(1) = (1 - 2)(1 - 3)q1(1) + 3
r(1) = 3q1(1) + 3

Substituting x = 2,

r(2) = (2 - 1)(2 - 3)q2(2) + 7
r(2) = -q2(2) + 7

Substituting x = 3,

r(3) = (3 - 1)(3 - 2)q3(3) + 13
r(3) = 2q3(3) + 13

Combining the equations:

Now, we can write the remainder r(x) as a polynomial of degree 2 as:

r(x) = Ax^2 + Bx + C

where A, B, C are constants.

Substituting the values of r(1), r(2), and r(3) in the above equation, we get:

3A + 3B + C = 3

Answer

P(x) when divided by x−1, x-2 and x−3 leaves remainder 3,7 and 13 respectively.
From polynomial-remainder theorem, P(1) = 3 and P(2) = 7 P(3) = 13
If the polynomial is divided by (x−1)(x−2)(x-3) then remainder must be of the form ax+b (degree of remainder is less than that of divisor)

⇒P(x) = Q(x)(x−1)(x−2)(x-3)+(ax² + bx + c), where Q(x) is some polynomial.
Substituting for x=1, x=2 and x=3:
P(1) = 3 = a+b+c
P(2) = 7 = 4a+2b+c
P(3) = 13 = 9a+3b+c
=> Subtract P(2) from P(1)
-3a - b = -4.............(1)
Subtract P(3) from P(2)
5a + b = 6..............(2)
From (1) and (2) Solving for a and b, we get
a = 1 and b = 1, c = 1
⇒Remainder = x² + x + 1

Answer

Correct answer is 'B'

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