Question

If $F_{1}$ and $F_{2}$ are irreducible factors of $x^{4}+x^{2}+1$ with real coefficients and $\frac{x^{3}-2 x^{2}+3 x-4}{x^{4}+x^{2}+1}=\frac{A x+B}{F_{1}}+\frac{C x+D}{F_{2}}$, then $A+B+C+D=$

• A. 2
• B. 1
• C. -3
• D. -4

Answer 1

Answer: (C)

We have,
$ x^{4}+x^{2}+1 =x^{4}+x^{2}+x^{2}-x^{2}+1 $
$=x^{4}+2 x^{2}+1-x^{2}=\left(x^{2}+1\right)^{2}-(x)^{2} $
$=\left(x^{2}+x+1\right)\left(x^{2}-x+1\right) $
Now,
$\frac{x^{3}-2 x^{2}+3 x-4}{x^{4}+x^{2}+1}=\frac{A x+B}{x^{2}+x+1}+\frac{C x+D}{x^{2}-x+1}$
$\Rightarrow \,x^{3}-2 x^{2}+3 x-4=(A x+B)$
$\left(x^{2}-x+1\right)+(C x+D)\left(x^{2}+x+1\right)$
$\Rightarrow \, x^{3}-2 x^{2}+3 x-4=(A+C) x^{3}$
$+x^{2}(B-A+C+D)+x(A-B+C+D)+(B+D)$
Comparing of all like terms
$ A +C=1 \,\,\,\,\,\,\,\dots(i)$
$ B-A+C+D =-2 $
$A-B+C+D =3 $
$ B+D =-4 \,\,\,\,\,\,\,\dots(ii)$
Adding Eqs. (i) and (ii), we get
$A+B+C+D=1+(-4)=-3$