Question

Let P(x) be any polynomial. When it is divided by (x - 2), (x - 3) and (x - 5), then the remainders are 5, 37 and 257 respectively. The remainder when P(x) is divided by (x-2)(x - 3)(x - 5) is

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Answer 1

Answer:

The remainder when P(x) is divided by (x-2)(x - 3)(x - 5) is   $26x^{2} - 98x + 97$.

Step-by-step explanation:

Let the polynomial be P(x)

P(x) mod (x-2) = 5

P(x) mod (x-3) = 37

P(x) mod (x-5) = 257

By Remainder theorem:

P(2) = 5

P(3) = 37

P(5) = 257

When P(x) is divided by (x-2)(x-3)(x-5), the remainder is of the form $ax^{2} + bx + c$.

Now,

Dividend = Divisor × Quotient + Remainder

$P(x) = (x-2) (x-3) (x-5) Q(x) + (ax^{2} +bx +c)$

Putting, P(2) = 5

$P(2) = a(2)^{2} + b(2) + c = 5$

       ⇒ $4a + 2b + c = 5$                     equation (1)

Putting, P(3) = 37

$P(3) = a(3)^{2} + b(3) + c = 37$

        ⇒ $9a + 3b + c = 37$                equation (2)

Putting, P(5) = 257

$P(5) = a(5)^{2} + b(5) + c = 257$

        ⇒ $25a + 5b +c =257$             equation (3)

Subtracting equation (1) from equation (2),

⇒ 9a + 3b + c  - ( 4a + 2b + c ) = 37 - 5

⇒ 9a + 3b + c - 4a - 2b - c = 32

⇒ 5a + b = 32                               equation (4)

Subtracting equation (2) from equation (3),

⇒ 25a + 5b + c  - ( 9a + 3b + c ) = 257 - 37

⇒ 25a + 5b + c - 9a - 3b - c = 220

⇒ 16a + 2b = 220                      equation (5)

Multiply equation (4) by 2 and subtract it from equation (5),

⇒ 16a + 2b - (10a + 2b) = 220 - 64

⇒ 16a + 2b - 10a -2b = 156

⇒  6a = 156

∴ a = 26

Putting value of a in equation (4),

⇒ 5 × 26 + b = 32

⇒ b = 32 - 130

∴ b =  - 98

Putting value of a and b in equation (1),

⇒ 4 × 26 + 2(-98) + c = 5

⇒ 104 - 196 + c = 5

⇒  c = 5 + 95

∴  c = 97

Therefore, Remainder is  $26x^{2} - 98x + 97$ .