Find the values of a, b and c

Question

The question:

Find values of $a,b,c.$ if $\displaystyle \frac{x^2+1}{x^2+3x+2} = \frac{a}{x+2}+\frac{bx+c}{x+1}$

My working so far:

http://i.imgur.com/VegifVa.jpg

How do I isolate $a$, $b$ and $c$?

Answer 1

For the right-hand side, let's obtain a common denominator. To that end, we obtain

$$\frac{a}{x+2}+\frac{bx+c}{x+1}=\frac{a(x+1)+(bx+c)(x+2)}{x^2+3x+1}=\frac{bx^2+(a+2b+c)x+(a+2c)}{x^2+3x+1}$$

Equating this last expression to $\frac{x^2+1}{x^2+3x+1}$ we see that

$$b=1$$

$$a+2b+c=0$$

and

$$a+2c=1$$

Can you finish from here?

SPOLIER ALERT: SCROLL OVER THE SHADED AREA TO REVEAL SOLUTION

We see that since $b=1$, the problem is reduced to solving a linear system of two equations for the two remaining unknowns, $a$ and $c$. We have $$a+c=-2$$and$$a+2c=1$$Upon solving, we find $a=-5$ and $c=3$.

Answer 2

Given $$\displaystyle \frac{x^2+1}{x^2+3x+2} = \frac{a}{x+2}+\frac{bx+c}{x+1} = \frac{a(x+1)+(bx+c)(x+2)}{x^2+3x+2}$$

So $$x^2+1 = bx^2+(a+2b+c)x+(a+2c)$$

Now equating Coefficients, we get $b=1$ and $(a+2b+c) = 0$ and $a+2c=1$

So Put $b=1$ in $a+2b+c=0\Rightarrow a+c=-2$ and above $a+2c=1$

so we get $c=3$ and $a=-5$

So we get $(a,b,c) = (-5,1,3)$