Question

Resolve $\frac{6x^4+11x^3+18x^2+14x+6}{(x+1)(x^2+x+1{)}^2}$ into partial fractions.

• A. $\frac{5}{x+1}+\frac{(x-1)}{(x^2+x+1)}+\frac{(3x+2)}{(x^2+x+1{)}^2}$
• B. $\frac{5}{x+1}-\frac{(x-1)}{(x^2+x+1)}+\frac{(3x+2)}{(x^2+x+1{)}^2}$
• C. $\frac{5}{x+1}+\frac{(x-1)}{(x^2+x+1)}-\frac{(3x+2)}{(x^2+x+1{)}^2}$
• D. $\frac{5}{x+1}+\frac{(x+1)}{(x^2+x+1)}+\frac{(3x+2)}{(x^2+x+1{)}^2}$

Answer 1

The correct option is A $\frac{5}{x+1}+\frac{(x-1)}{(x^2+x+1)}+\frac{(3x+2)}{(x^2+x+1{)}^2}$

The given fraction is a proper fraction
Let $\frac{6x^4+11x^3+18x^2+14x+6}{(x+1)(x^2+x+1{)}^2}=\frac{A}{x+1}+\frac{Bx+C}{x^2+x+1}+\frac{Dx+E}{(x^2+x+1{)}^2}$ ...(i)
$⟹6x^4+11x^3+18x^2+14x+6=A(x^2+x+1)+(Bx+C)(x+1)(x^2+x+1)$

$+(Dx+E)(x+1)$ ...(ii)
Putting $x+1=0$ or $x=-1$ in (ii), we obtain
$6(-1{)}^4+11(-1{)}^3+18(-1{)}^2+14(-1)+6=A(1-1+1{)}^2+0+0$
$⟹6-11+18-14+6=A$
$\therefore A=5$
Now puttitng $x^2+x+1=0$ or $x^2=-x-1$ in (ii), then
$6(-x-1{)}^2+11x(-x-1)+18(-x-1)+14x+6=0+0+D(-x-1)+Dx+Ex+E$
$⟹-5x^2-3x-6=-D+Ex+E$
$⟹-5(-x-1)-3x-6=-D+Ex+E$ ($\therefore x^2=-x-1$)
$⟹2x-1=(-D+E)+Ex$
on comparing, $E\ne 2$ and $-D+E=-1$ or $D=3$
Equating the coefficient of $x^4$ in (ii), we obtain
$6=A+B$
$⟹6=5+B$
$\therefore B=1$
Equzating the constant terms in (ii), we obtain
$6=A+B+E$
$⟹6=5+C+2$
$\therefore C=-1$
Substituting the value of a, B, C, D and E in (i), we get
$\frac{6x^4+11x^3+18x^2+4x+6}{(x+1)(x^2+x+1{)}^2}=\frac{5}{x+1}+\frac{(x-1)}{(x^2+x+1)}+\frac{(3x+2)}{(x^2+x+1{)}^2}$
which are the required partial fractions.
Hence, option 'A' is correct.