Bézout's Identity of polynomials?

Question

Let $P=X^3−7X+6$, $Q = 2X^2+ 5X − 3$ and $R = X^2 − 9 ∈\mathbb Q[X]$. What are $S$ and $T ∈\mathbb Q[X]$ such that $PS + QT = R$?

I have calculate the greatest common divisor of $P,Q,R$ are $(x+3)$, But this can only prove that the existence of $S,T$. So what should I do next?

Answer 1

.We know that the gcd is $x+3$, as you have found. Divide both polynomials by $(x+3)$ to get: $(x^2-3x+2)$ and $2x-1$ respectively. Now the gcd of these two polynomials is $1$. We do the following "reverse of the Euclidean algorithm": $x^2-3x+2 = 0.5x(2x-1) + (-2.5x + 2)$ and, $2x-1 = -0.8(-2.5x+2)+0.6$. Therefore,now letting $G=x^2-3x+2$ and $H=2x-1$, $$0.6=2x-1 +0.8(-2.5x+2) = 2x-1 + 0.8((x^2-3x+2) - 0.5x(2x-1)) = 0.8H + (1-0.4x)G.$$

Multiplying by $\frac{10}{6}(x+3)$, we get that $$1=\frac{25}{12}(P) + \frac{10}{6}(1-0.4x)(Q).$$ Now multiply both sides by $x^2-9$ to get the answer.

Answer 2

I like астон вілла олоф мэллбэрг’s answer fine, but when I divided your $P$ by $Q$, I got a remainder of $\frac34x+\frac94$, quotient $\frac12x-\frac54$. Thus \begin{align} \frac34x+\frac94&=P-\left(\frac12x-\frac54\right)Q\\ x+3&=\frac43P-\left(\frac23x-\frac53\right)Q\\ R&=\frac43(x-3)P + \left(-\frac23x^2+\frac{11}3x-5\right)Q\,. \end{align}