Find polynomials $u(x)$ and $v(x)$ such that $(2x + 3) \cdot u(x) + (x^2 + 1) \cdot v(x) = 1$.
I am trying to use the Euclidean algorithm for integers as a model, but I keep getting stuck. So far I have:
$x^2 + 1 = \frac{x}{2}(2x + 3) - \frac{3}{2}x + 1$ which leads to
$2x + 3 = -\frac{4}{3}(-\frac{3}{2}x + 1) + \frac{13}{3}$
and then I get stuck. I feel like there's something obvious that I'm doing wrong here, but I have no clue what it is. Can anyone help? Thanks!
On
I may be totally wrong but, as the question is set, there is an infinite number of polynomials $u(x)$ and $v(x)$ which can satisfy the relation $$(2x + 3) \, u(x) + (x^2 + 1) \, v(x) = 1$$
If $u(x)$ is of degree $n$, $v(x)$ would be of degree $n-1$ because of the multiplying terms. So, let us suppose $$u(x)=\sum_{i=0}^n a_ix^i$$ $$v(x)=\sum_{i=0}^{n-1} b_ix^i$$ Replacing and expanding, we have $$2x\sum_{i=0}^n a_ix^i+3\sum_{i=0}^n a_ix^i+x^2\sum_{i=0}^{n-1} b_ix^i+\sum_{i=0}^{n-1} b_ix^i=1$$ that is to say $$\sum_{i=0}^n 2a_ix^{i+1}+\sum_{i=0}^n 3a_ix^i+\sum_{i=0}^{n-1} b_ix^{i+2}+\sum_{i=0}^{n-1} b_ix^i=1$$ So, considering each degree, we shall have $$3a_0+b_0=1$$ $$2 a_0+3 a_1+b_1=0$$ $$2 a_1+3 a_2+b_0+b_2=0$$ $$2 a_2+3 a_3+b_1+b_3=0$$ $$2 a_3+3 a_4+b_2+b_4=0$$ and so on. From the above, you clearly see the pattern.
So, given coefficients $a_i$'s, you get coefficients $b_i$'s by successive elimination or , much better, using matrices.
Calculation is wrong.
$$x^2 + 1 = (\frac{x}{2} - \frac{3}{4})(2x + 3) + \frac{13}{4}$$ $$(2x + 3) = (\frac{8}{13}x +\frac{12}{13})\frac{13}{4} + 0$$
You need to find quotient for $(2x +3)|(x^2 + 1)$, and so on.