Question

The remainder when polynomial P(x) of degree 5 is divided by x+1 and x-1 is 1 and 2 respectively. Find the remainder when P(x) is divided by x2 -1.

Answer 1

,...............................

Answer 2

Answer: The remainder when P(x) is divided by x² -1 is (x+3)/2

Step-by-step explanation:

When P(x) is divided by x+1, remainder is 1.

∴ P(x) = (x + 1)q₁(x) + 1

Substitute x = -1 ,

P(-1)  = 1

When P(x) is divided by x-1, remainder is 2.

∴ P(x) = (x - 1)q₂(x) + 2

Substitute x = 1 ,

P(1)  = 2

Let the remainder be (ax + b) when P(x) is divided by (x² - 1).

∴ P(x) = (x² - 1)h(x) + ax + b  

Substitute x  = -1

P(-1) = -a + b

1 = b - a    ... eq (1)

P(x) = (x² - 1)h(x) + ax + b  

Substitute x  = 1

P(1) = a + b

2 = b + a     ... eq(2)

Adding eq(1) and eq(2) we get,

3 = 2b

b = 3/2

Substituting b = 3/2 in eq (2) we get,

2 = a + 3/2

a  = 1/2

The remainder is (x+3)/2.

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