Question

Find the Remainder when x^4+x^3–2x^2+x+1 is divided by x^2–1 (by Remainder theorem)
Is this possible yes or no ? Give reason for either answer?

Please don’t answer unnecessary i am stuck in this question if you don’t have the answer don’t answer.....

Answer 1

Answer:

Yes it is possible.

The remainder r(x) = 2x

Step-by-step explanation:

Let p(x) = x⁴ + x³-2x² + x + 1

d(x) = (x²-1) = (x-1)(x+1) = divisor

r(x) be the remainder

q(x) be the quotient.

Concepts

1) In polynomial division, degree of the remainder is always less that the degree of the divisor. Since the divisor d(x) is a 2 degree polynomial, the remainder will be a linear expression in this case.

2) The remainder when p(x) is divided by a factor of d(x) will be the same as the remainder when r(x) is divided by the same factor of d(x).(Proof is given in the last part of this answer)

Calculation

Since (x-1) and (x+1) are the factors of d(x), applying remainder theorem in addition to the above concepts, we get

p(1) = r(1)

or 1 + 1 -2 +1 +1 = a + b

or a + b = 2 .............(1)

And p(-1) = r(-1)

or 1 -1 - 2 - 1 + 1 = -a + b

or a - b = 2 ..............(2)

Solving (1) and (2) gives

a = 2, b = 0

Hence, the remainder r(x) = 2x

Proof of concept 2.

Let (x-a) be a factor of d(x). So d(a) = 0

We know that

dividen = divisor*quotient + remainder

⇒ p(x) = d(x)*q(x) + r(x)

Let us put x = a in this equation

⇒ p(a) = d(a)*q(a) + r(a)

⇒ p(a) = 0*q(a) +r(a)

⇒ p(a) = r(a)

Hence the remainders of p(x) and r(x) when divided by (x-a), a factor of d(x) is same.

I tried to explain with as much simplicity as possible. I hope this clear up your confusion.