Polynomials and common roots

Question

When dividing $f(x)$ by $g(x)$: $f(x)=g(x)Q(x)+R(x)$. How to find the quotient $Q(x)$ and the remainder $R(x)$? For example: $f(x)=\ 2x^4+13x^3+18x^2+x-4 \ $ , $g(x)=\ x^2+5x+2 \ $

At first $g(x)= (x+4.56)(x+0.44)$ then we use synthetic division : $Q(x)= \ 2x^2+3x-1 \ $ , $R(x)= \ 0.04x-1.98 \ $

how to show that $f(x)$ and $g(x)$ have no common roots?

I was wrong when I used synthetic division rather than long division so that, $Q(x)= \ 2x^2+3x-1 \ $ , $R(x)=-2 \ $

$f(x)= (2x^2+3x-1) g(x)\ $ $- 2 \ $

let $x=α\ $ is a root for $f(x)=0\ $ and $g(x)=0\ $ and substitute in the equation which proves α does not exist so they have no common roots.

Is that true?

Answer 1

Well $$(x + 4.56)(x + 0.44) = x^2 + 5x + 2.0064 \neq x^2 + 5x + 2 = g(x), $$ so no, anything using that factorization couldn't tell you anything about $g(x)$. Your $Q(x)$ looks OK, but your $R(x)$ certainly doesn't satisfy $f(x) = Q(x)g(x) + R(x)$. Now, once you use $g(x)$ to get the correct $R(x)$ you will have an expression of the form $$ R(x) = f(x) - Q(x)g(x) $$ and so if some factor divides both $f(x)$ and $g(x)$, what can you say about $R(x)$? Can you see why this shows $f$ and $g$ have no common roots?

Answer 2

Hint:

If two polynomials have common roots, they have a common factor (which is the product of the binomials $z-r_i$).

This common factor is their $\gcd$, for which the Euclidean algorithm can be used (divide $p$ by $q$; if there is a remainder, let $r$, divide $q$ by $r$, and so on).