Problem: Determine all values for [imath]a[/imath] such that the polynomial [imath]p(x)=x^3+2x^2-x-2[/imath] and [imath]g(x)=x^3+2x+a[/imath] are given non-trivial greatest common divisors and specify these.
I believe the structure of the argument should be of the following form. To find the values of [imath]a[/imath] such that the polynomials [imath]p(x)[/imath] and [imath]g(x)[/imath] we will have non-trivial greatest common divisors we use the Euclidean algorithm. The greatest common divisor of two polynomials divides both polynomials without remainder. If the GCD is non-trivial, it means it is not just a constant multiple of one of the polynomials.
We first calculate [imath]p(x)/g(x)[/imath]. By the division-algorithm theorem there exist some polynomials [imath]q_1(x)[/imath] and [imath]r_1(x)[/imath] such that
[math]p(x)=g(x)q_1(x)+r_1(x) , deg(r_1)<deg(g)=3[/math][math]g(x)=r_1(x)q_2(x)+r_2(x), deg(r_2)<deg(r_1)[/math][math]r_1(x)=r_2(x)q_3(x) + 0, deg(r_3)=0[/math]
But since [imath]gcd(p(x),g(x))=gcd(p(x)-g(x)q_1(x),g(x))=gcd(r_1(x),g(x))=gcd(r_1(x),g(x)-r_1(x)q_2(x))=gcd(r_1(x),r_2(x)))=gcd(r_1(x)-r_2(x)q_3(x),r_2(x))=gcd(0,r_2(x))=r_2(x)[/imath].
We would conclude that any common divisor for [imath]p(x)[/imath] and [imath]g(x)[/imath] must be the polynomial [imath]r_2(x)[/imath] which depends on the constant [imath]a[/imath]. Since the polynomial [imath]r_2(x)[/imath] is the greatest common divisor for [imath]f(x)[/imath]and [imath]g(x)[/imath]and is not a constant polynomial, so we must restrict [imath]a[/imath] such that [imath]r_2=/0[/imath] and [imath]r_3(x)=0[/imath].
This is my computation attempt.
[math]p(x)=x^3+2x^2-x-2=1(x^3+2x+a)+(2x^2-3x-(2+a))=q_1(x)g(x)+r_1(x)[/math]
for [imath]q_1(x)=1[/imath] and [imath]r_1(x)= (2x^{2}-3x-(2+a))[/imath]
[math]g(x)=r_1(x)(\frac{x}{2}+\frac{3}{4})+(\frac{1}{4}(x(21+2a)+(6+7a))[/math]
for [imath]q_2(x)=\frac{x}{2}+\frac{3}{4}[/imath] and [imath]r_2(x)=\frac{1}{4}(x(21+2a)+(6+7a)[/imath]
[math]r_1(x)=r_2(x)(\frac{8x}{21+2a}-\frac{4(62a+111)}{(2a+21)^2})+ (\frac{4(62a+111)(7a+6)}{(2a+21)^2}-(2+a))=r_2(x)q_3(x)+r_3(x)[/math]
for [imath]q_3(x)=\frac{8x}{21+2a}-\frac{4(62a+111)}{(2a+21)^2}[/imath] and [imath]r_3(x)=(\frac{4(62a+111)(7a+6)}{(2a+21)^2}-(2+a))[/imath]
Here I'm stuck since to determine the condition such that [imath]r_3(x)=0[/imath], you must solve a third degree equation, all I conclude from the above would be that [imath]a\neq-21/2[/imath]. I'm pretty sure I've made a misstake in my reasoning and any advice would be appreciated.
I believe the structure of the argument should be of the following form. To find the values of [imath]a[/imath] such that the polynomials [imath]p(x)[/imath] and [imath]g(x)[/imath] we will have non-trivial greatest common divisors we use the Euclidean algorithm. The greatest common divisor of two polynomials divides both polynomials without remainder. If the GCD is non-trivial, it means it is not just a constant multiple of one of the polynomials.
We first calculate [imath]p(x)/g(x)[/imath]. By the division-algorithm theorem there exist some polynomials [imath]q_1(x)[/imath] and [imath]r_1(x)[/imath] such that
[math]p(x)=g(x)q_1(x)+r_1(x) , deg(r_1)<deg(g)=3[/math][math]g(x)=r_1(x)q_2(x)+r_2(x), deg(r_2)<deg(r_1)[/math][math]r_1(x)=r_2(x)q_3(x) + 0, deg(r_3)=0[/math]
But since [imath]gcd(p(x),g(x))=gcd(p(x)-g(x)q_1(x),g(x))=gcd(r_1(x),g(x))=gcd(r_1(x),g(x)-r_1(x)q_2(x))=gcd(r_1(x),r_2(x)))=gcd(r_1(x)-r_2(x)q_3(x),r_2(x))=gcd(0,r_2(x))=r_2(x)[/imath].
We would conclude that any common divisor for [imath]p(x)[/imath] and [imath]g(x)[/imath] must be the polynomial [imath]r_2(x)[/imath] which depends on the constant [imath]a[/imath]. Since the polynomial [imath]r_2(x)[/imath] is the greatest common divisor for [imath]f(x)[/imath]and [imath]g(x)[/imath]and is not a constant polynomial, so we must restrict [imath]a[/imath] such that [imath]r_2=/0[/imath] and [imath]r_3(x)=0[/imath].
This is my computation attempt.
[math]p(x)=x^3+2x^2-x-2=1(x^3+2x+a)+(2x^2-3x-(2+a))=q_1(x)g(x)+r_1(x)[/math]
for [imath]q_1(x)=1[/imath] and [imath]r_1(x)= (2x^{2}-3x-(2+a))[/imath]
[math]g(x)=r_1(x)(\frac{x}{2}+\frac{3}{4})+(\frac{1}{4}(x(21+2a)+(6+7a))[/math]
for [imath]q_2(x)=\frac{x}{2}+\frac{3}{4}[/imath] and [imath]r_2(x)=\frac{1}{4}(x(21+2a)+(6+7a)[/imath]
[math]r_1(x)=r_2(x)(\frac{8x}{21+2a}-\frac{4(62a+111)}{(2a+21)^2})+ (\frac{4(62a+111)(7a+6)}{(2a+21)^2}-(2+a))=r_2(x)q_3(x)+r_3(x)[/math]
for [imath]q_3(x)=\frac{8x}{21+2a}-\frac{4(62a+111)}{(2a+21)^2}[/imath] and [imath]r_3(x)=(\frac{4(62a+111)(7a+6)}{(2a+21)^2}-(2+a))[/imath]
Here I'm stuck since to determine the condition such that [imath]r_3(x)=0[/imath], you must solve a third degree equation, all I conclude from the above would be that [imath]a\neq-21/2[/imath]. I'm pretty sure I've made a misstake in my reasoning and any advice would be appreciated.
