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This is probably elementary , but I cannot figure it out . My book says using long division, dividing $2x^2 +4x -3$ by $2x-1$ gives a quotient of $x+2$ and a remainder of $x-1$ .But when I do long division I get a quotient of $x+\frac52$ and a remainder of $-\frac12$ ...what is going on here??? How is the book answer obtained?

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    $\begingroup$ Book is obviously wrong. degree of remainder will be less than divisor. $\endgroup$
    – akhmeteni
    Sep 27, 2017 at 6:00
  • $\begingroup$ Write $2\left(2x^2+4x-3\right)=(2x-1)(2x+5)-1$ over the integers. It is then clear that $g c d\left(2x^2+4x-3,2x-1\right)=1$ $\endgroup$
    – Lozenges
    Sep 27, 2017 at 6:28

2 Answers 2

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Book version:

$$(2x-1)(x+2) + (x-1) = 2x^2+4x-3$$

Your version: $$(2x-1)(x+2.5)-0.5=2x^2+4x-3$$

If we adopt the convention that the remainder must have a smaller degree, then your answer is correct unless there are restriction such as the coefficient of the polynomial has to be integered value.

Edit:

$$2x^2+4x-3=(2x-1)(x+B)+(Cx+D)$$

Let's determine $B, C,D \in \mathbb{Z}$.

$$2x^2+4x-3=2x^2+(2B-1)x-B+(Cx+D)$$

We can see that $2B-1$ must be an odd number. let's pick $2B-1$ to be $3$, hence $B=2$.

$$2x^2+4x-3=2x^2+3x-2+(Cx+D)$$

Hence $C=1$ and $D=-1$.

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  • $\begingroup$ Yes ,thanks i can see both answers are correct , but how this they get their answer using long division? $\endgroup$
    – herashefat
    Sep 27, 2017 at 6:10
  • $\begingroup$ Another thing I want to add is that this computation comes up in finding gcd of polynomials so we cannot have fractions..... $\endgroup$
    – herashefat
    Sep 27, 2017 at 6:14
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If you restrain the division to $\mathbb Z[X]$ (i.e. with only integer coefficients) you will get the book version.

$\begin{array}{r|r} 2x^2+4x-3 & 2x-1\\\hline -(2x^2-x) & x+2\\ 5x-3\\ -(4x-2)\\ x-1\end{array}$

So first there is $x$ times $2x-1$, you substract $2x^2-x$ to get $5x-3$.

Now in $5x-3$ it goes only $2$ times $2x-1$ if we don't want fractions, and $3$ is too large to keep the leading coefficient positive.

While if you allow dividing in $\mathbb R[X]$ then you get your version with a $\frac 52$ and a remainder with degree strictly inferior to the divisor.

Although when fractions appear in the division of $P$ by $Q$ in $\mathbb Z[x]$ there is another convention which is to say that $\lambda P=QD+R$ with $\deg R<\deg Q$.

In this case we have $2(5x-3)=5(2x-1)-1$ instead of $(5x-3)=2(2x-1)+(x-1)$

And we get the integer division $2(2x^2+4x-3)=(2x-1)(2x+5)-1$

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