Polynomial/Euclidean division/6x^3+x+1 over 3x^2+2x-4/Example

We want to apply the Euclidean division (over ${}\mathbb {Q}$)

$P=6X^{3}+X+1{\text{ divided by }}T=3X^{2}+2X-4.$

So we want to divide a polynomial of degree ${}3$ by a polynomial of degree ${}2$, hence the quotient and also the remainder have (at most) degree ${}1$. For the first step, we ask with which term we have to multiply ${}T$ to achieve that the product and ${}P$ have the same leading term. This is ${}2X$. The product is

${}2X{\left(3X^{2}+2X-4\right)}=6X^{3}+4X^{2}-8X\,.$

The difference between ${}P$ and this product is

${}6X^{3}+X+1-{\left(6X^{3}+4X^{2}-8X\right)}=-4X^{2}+9X+1\,.$

We continue the division by ${}T$ with this polynomial, which we call ${}P'$. In order to get coincidence with the leading coefficient we have to multiply ${}T$ with ${}{\frac {-4}{3}}$. This yields

${}-{\frac {4}{3}}T=-{\frac {4}{3}}{\left(3X^{2}+2X-4\right)}=-4X^{2}-{\frac {8}{3}}X+{\frac {16}{3}}\,.$

The difference between this and ${}P'$ is therefore

${}-4X^{2}+9X+1-{\left(-4X^{2}-{\frac {8}{3}}X+{\frac {16}{3}}\right)}={\frac {35}{3}}X-{\frac {13}{3}}\,.$

This is the remainder and altogether we get

${}6X^{3}+X+1={\left(3X^{2}+2X-4\right)}{\left(2X-{\frac {4}{3}}\right)}+{\frac {35}{3}}X-{\frac {13}{3}}\,.$