What's the best way to solve $1 = A(x^2+1) + (Bx+C)(x+1)$

Question

What's the best way to solve $$1 = A(x^2+1) + (Bx+C)(x+1)$$

I let $x=-1$ and got $A=\frac{1}{2}$ But what sub is ideal to find B&C

This gets messy quick, I think.

Instead, I started over, and I grouped like terms

$$1 = Ax^2+A+Bx^2+Bx+Cx+C$$ This led to $$A+B=0$$ $$B+C=0$$ $$A+C=1$$ This led to $B=-\frac{1}{2}$

$C=\frac{1}{2}$

and solved using my A=1/2 But, I did not like that I had to mix techniques.

Answer 1

At each step take advantage of your previous findings.$$1 = A(x^2+1) + (Bx+C)(x+1)$$

$$x=-1 \implies A=\frac {1}{2}$$ $$ x=0 \implies A+C=1 $$

Thus $$C =1/2$$ $$x=1 \implies 2A+2(B+C)=1$$ Thus $$B+C=0 \implies B=-\frac {1}{2}$$

Answer 2

Grouping like terms is the best way to solve for the coefficients of any polynomial. Like @Paul said, you did it exactly right.

First, we group like terms together: $$A(x^2 + 1) + (Bx + C)(x+1) = 1$$ $$(A+B)x^2 + (B+C)x + A+C=1 + 0x + 0x^2$$ Second, we solve for the coefficients: $$A+B=0\implies A=-B$$ $$B+C=0 \implies C = -B = A$$ $$A + C = 1 \implies A+A = 2A=1 \implies A = \frac{1}{2}.$$ $$B=-A=-\frac{1}{2}$$ $$C = A = \frac{1}{2}.$$ You used mixed techniques, which is fine and mathematically valid; however, it was unnecessary here.