Is this method to calculate gcd of two polynomials correct?

Question

I have the following two polynomials: $$a(x) = x^3+4x^2+x-6,b(x) = x^5-6x+5$$

We observe that $a(x) = (x-1)(x+2)(x+3).$ We observe that $b(1)=0$ and $b(-2),b(-3)$ are not equal to $0.$ Thus $\gcd(a,b) = x-1.$

Answer 1

Yes, this is correct. More generally: if you have two polynomials $p(x)$ and $q(x)$, if$$\{a_1,\ldots,a_k\}=\{\text{roots of }p(x)\}\cup\{\text{roots of }q(x)\},$$and if $\alpha_1,\beta_1,\ldots,\alpha_k,\beta_k\in\mathbb{Z}^+$ are such that$$p(x)=(x-a_1)^{\alpha_1}\cdots(x-a_k)^{\alpha_k}\text{ and }q(x)=(x-a_1)^{\beta_1}\cdots(x-a_k)^{\beta_k},$$then$$\gcd\bigl(p(x),q(x)\bigr)=(x-a_1)^{\min\{\alpha_1,\beta_1\}}\cdots(x-a_k)^{\min\{\alpha_k,\beta_k\}}.$$In your case, your polynomials have one and only one common root, which is a simple root of one of them.

Answer 2

$$ \left( x^{5} - 6 x + 5 \right) $$

$$ \left( x^{3} + 4 x^{2} + x - 6 \right) $$

$$ \left( x^{5} - 6 x + 5 \right) = \left( x^{3} + 4 x^{2} + x - 6 \right) \cdot \color{magenta}{ \left( x^{2} - 4 x + 15 \right) } + \left( - 50 x^{2} - 45 x + 95 \right) $$ $$ \left( x^{3} + 4 x^{2} + x - 6 \right) = \left( - 50 x^{2} - 45 x + 95 \right) \cdot \color{magenta}{ \left( \frac{ - 10 x - 31 }{ 500 } \right) } + \left( \frac{ 11 x - 11 }{ 100 } \right) $$ $$ \left( - 50 x^{2} - 45 x + 95 \right) = \left( \frac{ 11 x - 11 }{ 100 } \right) \cdot \color{magenta}{ \left( \frac{ - 5000 x - 9500 }{ 11 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( x^{2} - 4 x + 15 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{2} - 4 x + 15 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - 10 x - 31 }{ 500 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 10 x^{3} + 9 x^{2} - 26 x + 35 }{ 500 } \right) }{ \left( \frac{ - 10 x - 31 }{ 500 } \right) } $$ $$ \color{magenta}{ \left( \frac{ - 5000 x - 9500 }{ 11 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 100 x^{4} + 100 x^{3} + 100 x^{2} + 100 x - 500 }{ 11 } \right) }{ \left( \frac{ 100 x^{2} + 500 x + 600 }{ 11 } \right) } $$ $$ \left( x^{4} + x^{3} + x^{2} + x - 5 \right) \left( \frac{ - 10 x - 31 }{ 55 } \right) - \left( x^{2} + 5 x + 6 \right) \left( \frac{ - 10 x^{3} + 9 x^{2} - 26 x + 35 }{ 55 } \right) = \left( -1 \right) $$ $$ \left( x^{5} - 6 x + 5 \right) = \left( x^{4} + x^{3} + x^{2} + x - 5 \right) \cdot \color{magenta}{ \left( x - 1 \right) } + \left( 0 \right) $$ $$ \left( x^{3} + 4 x^{2} + x - 6 \right) = \left( x^{2} + 5 x + 6 \right) \cdot \color{magenta}{ \left( x - 1 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x - 1 \right) } $$ $$ \left( x^{5} - 6 x + 5 \right) \left( \frac{ - 10 x - 31 }{ 55 } \right) - \left( x^{3} + 4 x^{2} + x - 6 \right) \left( \frac{ - 10 x^{3} + 9 x^{2} - 26 x + 35 }{ 55 } \right) = \left( - x + 1 \right) $$

.........