How do you use the remainder theorem to find the remainder for each division #(x^2-2)div(x-1)#?

Question

2509 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-11T04:43:54

Quotient # (x+1)# & Remainder #3#

          x          1
      ………………………………………………………

x-1 | x^2 0 2
| x^2 -x
|__
0 x 2
x -1
_
0 3 (Remainder)

Answer 2 · 2017-10-11T07:03:06

The remainder is #-1#. See explanation.

The remainder theorem says that:

For a polynomial #P(x)# the remainder of division #P(x)-:(x-a)# is equal to #P(a)#.

So instead of using the long division to calculate the remainder only you can substitute #a# to #P(x)#. Here we have:

#a=1#, #P(x)=x^2-2#, so:

#P(1)=1^2-2=1-2=-1#

The remainder of dividision #(x^2-2)-:(x-1)# is #-1#