Find the remainder of a division

[pixel01]

Hi all!

Could anyone help telling me the way to find the remainder of the following divisions:

1. (x^2006+x^1996+x^1981+x+1):(x^2-1)

2. (x2+x3+x5+1) : [(x-1)(x-2)]

Thanks

 

[HallsofIvy]

Saying that "P(x) divided by U(x) is equal to quotient Q(x) with remainder R(x) means that P(x)= U(x)*Q(x)+ R(x)." Of course, R(x) has lower degree than U(x) so that, for example, if U is linear, then R(x)= r, a number. In both of your examples, U(x) is quadratic so R(x) is linear. It is also true that, if U(a)= 0 then P(a)= R(a).

So: for the first problem, Q(x)= x²- 1 which has zeroes 1 and -1. You know that the remainder is a linear function, R(x)= ax+ b, such that R(1)= 1^2006+1^1996+1^1981+1+1= ? and R(-1)= (-1)^2006+(-1)^1996+(-1)^1981+(-1)+1= ?. Two points are sufficient to determine a and b. Same idea for the second problem.

 

[pixel01]

...
So: for the first problem, Q(x)= x²- 1 which has zeroes 1 and -1. You know that the remainder is a linear function, R(x)= ax+ b, such that R(1)= 1^2006+1^1996+1^1981+1+1= ? and R(-1)= (-1)^2006+(-1)^1996+(-1)^1981+(-1)+1= ?. Two points are sufficient to determine a and b. Same idea for the second problem.

There may be a mistake in your reasoning. U(x) equals x^2-1 which has two roots at 1 and -1, not Q(x).

 

[D H]

There is a typo in his reasoning, not a mistake. Substitute Q(x) with U(x) in Hall's write-up. It doesn't matter what Q(x) is. It is just some polynomial. If U(x) is zero for some particular value of x, the product Q(x)*U(x) will also be zero at that value of x.

For a polynomial divided by a quadratic, you know that the remainder R(x) must be of the form ax+b. Since P(x) = Q(x)*U(x) + R(x), P(x) and R(X) must be equal at the zeros of U(x). Evaluating P(x) at the the zeros of U(x) provides the information needed to deduce the form of R(x).

 

[pixel01]

Oh, now I've got to the point:
P(x) - R(x)=U(x).Q(x)
At x=1 and x=-1, the left-hand side equals zero ..

Thank you HallsofIvy and DH.