I have been thinking about this. Lets say we take 7 divided by 3, we know the remainder is 1. However, if we let x=7 and x-4=3, and we take x/(x-4), after performing long division the remainder is 4. Why is it not 1?

Corbin Bradford 2022-09-22 Answered
Remainder and long division
I have been thinking about this. Lets say we take 7 divided by 3, we know the remainder is 1. However, if we let x=7 and x-4=3, and we take x/(x-4), after performing long division the remainder is 4. Why is it not 1?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

ahem37
Answered 2022-09-23 Author has 15 answers
Because the quotients are different:
7 = 2 3 + 1
x = 1 ( x 4 ) + 4
When x = 7 we get
7 = 1 3 + 4
which is correct, of course.
Did you like this example?
Subscribe for all access
Janet Hart
Answered 2022-09-24 Author has 2 answers
If someone handed you x and x 4, why would you assume that they are 7 and 3?
In general, we know that x x 4 = 1 + 4 x 4
For x = 7, it is the same that 7 7 4 = 1 + 4 7 4 , just like any arbitrary x
However, it just so happens that 1 + 4 7 4 can be rewritten as 2 + 1 7 4
The polynomial remainder is still 4
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2021-11-16
Solve: 5|x3|+2>4
The answer has the form:
a) (A,B)
b) (,A)(B,)
State your solution using interval notation, using whole numbers, proper fractions or improper fractions, like above: ?
asked 2022-07-21
Showing 1 2 3 2 p ( x + y + z ) p 1 x p y + z + y p x + z + z p y + x with p > 1, and x , y , z positive
By Jensen I got taking p 1 = p 2 = p 3 = 1 3 , (say x = x 1 , y = x 2 , z = x 3 )
( p k x k ) p p k x p which means;
3 1 p ( x + y + z ) p x p + y p + z p
and if I assume wlog x y z then 1 y + z 1 x + z 1 x + y
so these 2 sequences are similarly ordered, and by rearrangement inequality the rest follows ?
I should use power means, but any other solution is also appreciated.
asked 2022-03-16
Common denominator with radicals
Can you make these 2 fractions into 1?
292x2x92x
I thought you could make them into 2x+1892x
asked 2022-06-15
Prove inequality by induction n N , n 2     :     ( 1 + 1 n ) n < k = 0 n 1 k ! < 3
I have the following statement to prove:
n N , n 2     :     ( 1 + 1 n ) n < k = 0 n 1 k ! < 3
What I have already proven is that
( 1 + 1 n ) n < ( 1 + 1 n + 1 ) n + 1
I have started, as usual by induction with n = 2. Then I went on to say
( 1 + 1 n + 1 ) ( 1 + 1 n + 1 ) n < k = 0 n 1 k ! + 1 ( n + 1 ) ! < 3
But this is where I seem to get stuck. I cannot use the induction hypothesis, since the denominator on the left side is now n + 1 instead of n. I assume, I have to use the other inequality, which I have already proven to be true, but I do not know how. Any hints?
asked 2021-12-30
Write 78% as a fraction
asked 2021-12-17
How do you integrate 1x2
asked 2022-08-11
Find the value of 1 1 2 + 1 + 1 2 2 + 2 + 1 3 2 + 3 + 1 4 2 + 4 + + + 1 2008 2 + 2008
My attempt:
I tried to think of a better way to handle:
1 n 2 + n
Then I got (doesn't work only on 1 1 2 + 1 ):
n 1 n 3 n
By putting the values in, I got:
1 2 + 2 1 2 3 2 + 3 1 3 3 3 + 4 1 4 3 4 + + 2008 1 2008 3 2008
It's still doesn't make sense. Is there another way of solving this question? Can I have a hint or a guide?