1. ## Hcf polynomials

Hi there,
(a)
let $f(X) = 2X^4 + 5X^3 + 8X^2 + 7X + 4$ and $g(X) = 2X^2 + 3X + 3$
the HCF(f, g) = 1.

but is there a special way to do this kind of question... or do i have to compute manually by long division?

(b)

now let $\alpha\epsilon C$ satisfy $f(\alpha) = 0$.
Find a,b,c,d $\epsilon Q$ such that :

$1/g(\alpha) = a + b\alpha + c\alpha^2 + d\alpha^3$

any clue on how to go about with this one please? ill try it out for myself once i have an idea on what route to follow?

thnx a great deal guys

2. If HCF(f,g)=1 then exists two polynomials $h_1,h_2$ such as
$h_1f+h_2g=1$

In this case, applying Euclid's algorithm, we have:
$f=(X^2+X+1)g+X+1$ (1)
$g=(2X+1)(X+1)+2$
so, HCF(f,g)=1.

Then, $\displaystyle X+1=\frac{g-2}{2X+1}$
Plug in (1) and we have $(2X+1)f-(2x^3+3X^2+3X+2)g=-2$
Divide by -2: $\left(-X-\frac{1}{2}\right)f+\left(X^3+\frac{3}{2}X^2+\frac {3}{2}X+1\right)g=1$

Now, for $x=\alpha$ we have $\left(\alpha^3+\frac{3}{2}\alpha^2+\frac{3}{2}\alp ha+1\right)g(\alpha)=1\Leftrightarrow\frac{1}{g(\a lpha)}=\alpha^3+\frac{3}{2}\alpha^2+\frac{3}{2}\al pha+1$
Then $a=1,b=\frac{3}{2},c=\frac{3}{2},d=1$