# FInd gcd of two polynomials using Euclidean Algorithm

Let $f(x)=2x^4 +3x^3 −19x^2 −28x+ 6$ and $g(x)=x^3 + 2x^2 -9x -18$ be polynomials in $\mathbb Q[x]$. Use the Euclidean Algorithm to determine the gcd in $\mathbb Q[x]$.

So far, I have the following:

$2x^4 + 3x^3 -19x^2 -28x +6 = 2x(x^3 +2x^2 -9x -18)+(-x^3 -x^2 +8x +6)$

$x^3 +2x^2 -9x -18 = 1(-x^3 -x^2 +8x +6)+(3x^2 -17x -24)$

$-x^3 -x^2 +8x +6 = [(-1/3)x](3x^2 -17x -24)+[(-20/3)x^2 +6]$

This is where I am stuck…

I'm not even sure if I am doing each step right. A little point in the right direction would be greatly appreciated. Thanks.

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Note: as a first step you want to find $f(x)=q(x)g(x)+r(x)$ where the degree of $r(x)$ is less than the degree of $g(x)$ - so you want to keep dividing through until you have a remainder which is quadratic (or less). Your basic organisation is OK, but you should be reducing the degree of the remainder at each stage - you aren't completing all the divisions. So $-x^3-x^2+8x+6$ in the first line is a cubic, and you can get it down to a quadratic. –  Mark Bennet Apr 8 '14 at 22:06
Use long division method!! –  L16H7 Apr 11 at 10:11

## 2 Answers

Let a(x) and b(x) be your polynomials a(x)=q(x)*b(x)+r(x) .where r(x) is your remainder and q(x) your quotient dont forget deg(r(x))< de(b(x)) repeat untill your alorithm has a remainter with degree no greater than 1.!!

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Given any two polynomials, if we keep using the euclidean algorithm then eventually we will always get to a constant. So, is the gcd always a constant? –  Al Jebr May 15 at 21:15
@AlJebr IF you keep using euclidean algorithm you will always get zero the last non zero remainder is the gcd(alsothe gcd must be a monic polynomial ) .Thats just all about it.It isnt sure if the last non zero remainder will be a constant or not. –  Manolis Lyviakis May 26 at 22:30

In your last step, the remainder should have degree strictly less than the degree of the polynomial you are dividing by, so the division in that step isn't finished. The remainder should have degree no greater than 1, since you are dividing by a polynomial of degree 2.

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Given any two polynomials, if we keep using the euclidean algorithm then eventually we will always get to a constant. So, is the gcd always a constant? –  Al Jebr May 15 at 21:16