Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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 2

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.!!

share|improve this answer

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.