Expand the quotient by partial fractions.
-$$\frac { x + 4 } { x ^ { 2 } + 2 x + 1 }$$
A) $\frac { 3 } { x + 1 } + \frac { 1 } { ( x + 1 ) ^ { 2 } }$
B) $\frac { 1 } { x + 1 } + \frac { 4 } { x + 4 }$
C) $\frac { 1 } { x + 1 } + \frac { 3 } { ( x + 1 ) ^ { 2 } }$
D) $\frac { 1 } { x + 1 } + \frac { - 3 } { ( x + 1 ) ^ { 2 } }$

Question

Quizplus · Accepted Answer

We need to expand the given fraction as a sum of two simpler fractions with denominators of the form $x+1$ and $(x+1)^2,$ respectively:
$$\frac{x+4}{x^2+2x+1}=\frac{A}{x+1}+\frac{B}{(x+1)^2}$$Multiplying both sides by the denominator on the left, we get:
$$x+4=A(x+1)+B$$Expanding the right-hand side gives $Ax+A+B,$ so $A=3$ and $B=1.$ Therefore,
$$\frac{x+4}{x^2+2x+1}=\frac{3}{x+1}+\frac{1}{(x+1)^2}=\boxed{	extbf{(C)}}.$$

Expand the Quotient by Partial Fractions x+4x2+2x+1\frac { x + 4 } { x ^ { 2 } + 2 x + 1 }x2+2x+1x+4​

Expand the Quotient by Partial Fractions $\frac { x + 4 } { x ^ { 2 } + 2 x + 1 }$