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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1$\begingroup$ Book is obviously wrong. degree of remainder will be less than divisor. $\endgroup$– akhmeteniSep 27, 2017 at 6:00
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$\begingroup$ Write $2\left(2x^2+4x-3\right)=(2x-1)(2x+5)-1$ over the integers. It is then clear that $g c d\left(2x^2+4x-3,2x-1\right)=1$ $\endgroup$– LozengesSep 27, 2017 at 6:28
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2 Answers
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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 $D=-1$.
answered Sep 27, 2017 at 6:02
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$\begingroup$ Yes ,thanks i can see both answers are correct , but how this they get their answer using long division? $\endgroup$ Sep 27, 2017 at 6:10
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$\begingroup$ Another thing I want to add is that this computation comes up in finding gcd of polynomials so we cannot have fractions..... $\endgroup$ 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 $2(2x^2+4x-3)=(2x-1)(2x+5)-1$
