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Simple Remainder Therom

Why is it that when you divide a a polynomial by a quadratic divisor that it's remainder is linear e.g. 4x + 7 and not just a constant number e.g. 8?

e.g. 2x2+4x+5x21\dfrac{2x^2 + 4x + 5}{x^2 - 1}

When comparing coefficients, could someone briefly explain how I would 'deduce' that f(x)=A(x21)+Bx+C1f(x) = A(x^2 - 1) + \dfrac{Bx + C}{1} for this particular question.

I'm doing some basic comparing coefficients and was just curious as the textbook doesn't really explain this (it probably shouldn't either since this is basic high-school maths)
(edited 10 years ago)
Original post by Inevitable
Why is it that when you divide a a polynomial by a quadratic divisor that it's remainder is linear?

e.g. 2x2+4x+5x21\dfrac{2x^2 + 4x + 5}{x^2 - 1}

When comparing coefficients, could someone briefly explain how I would 'deduce' that f(x)=A(x21)=Bx+C1f(x) = A(x^2 - 1) = \dfrac{Bx + C}{1} for this particular question.

I'm doing some basic comparing coefficients and was just curious as the textbook doesn't really explain this (it probably shouldn't either since this is basic high-school maths)


The remainder has to be linear (or a constant) as the degree necessarily has to be less than the degree of the divisor. The degree of the divisor here is 2 so the degree of the remainder has to be 1 or 0.

I don't understand the second part of your question - have you made a mistake?
Reply 2
Original post by Inevitable
Why is it that when you divide a a polynomial by a quadratic divisor that it's remainder is linear e.g. 4x + 7 and not just a constant number e.g. 8?

e.g. 2x2+4x+5x21\dfrac{2x^2 + 4x + 5}{x^2 - 1}

When comparing coefficients, could someone briefly explain how I would 'deduce' that f(x)=A(x21)=Bx+C1f(x) = A(x^2 - 1) = \dfrac{Bx + C}{1} for this particular question.

I'm doing some basic comparing coefficients and was just curious as the textbook doesn't really explain this (it probably shouldn't either since this is basic high-school maths)



Because it can be

It is not necessarily but, as I say, it can be


A+Bx+Cx2+3x1A + \dfrac{Bx+C}{x^2+3x-1}

and

A+Cx2+3x1A + \dfrac{C}{x^2+3x-1}


Both give a quadratic divided by x2+3x1x^2+3x-1 when added together

So, when splitting the fraction back up into it's constituent parts we have to accept that it could be the former rather than the latter


If it turns out to be the latter then calculation with the former would lead to B = 0
(edited 10 years ago)
Reply 3
Original post by TenOfThem
Because it can be

It is not necessarily but, as I say, it can be


A+Bx+Cx2+3x1A + \dfrac{Bx+C}{x^2+3x-1}

and

A+Cx2+3x1A + \dfrac{C}{x^2+3x-1}


Both give a quadratic divided by x2+3x1x^2+3x-1 when added together

So, when splitting the fraction back up into it's constituent parts we have to accept that it could be the former rather than the latter


If it turns out to be the latter then calculation with the former would lead to B = 0


I see (so we always assume the remainder is linear i.e. (Bx+C(Bx + C) and if it is a constant, then we will figure it out when comparing coefficients as B will equal 0?

Although, why are we multiplying the divisor by a constant A to get the quotient?
(edited 10 years ago)
Reply 4
Original post by Inevitable


Although, why are we multiplying the divisor by a constant A to get the quotient?


You know the answer to that

What would you do if I asked you to convert 3253\frac{2}{5} to a top heavy fraction
Reply 5
Original post by TenOfThem
You know the answer to that

What would you do if I asked you to convert 3253\frac{2}{5} to a top heavy fraction


I would multiply 3 by 5, add 2 to get 17/5.

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