Partial fractions Questions in English

(D) Given the partial fraction decomposition: $\frac{x^2-x+1}{(x^2+1)(x^2+x+1)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+x+1}$
Multiplying both sides by $(x^2+1)(x^2+x+1)$,we get:
$x^2-x+1 = (Ax+B)(x^2+x+1) + (Cx+D)(x^2+1)$
$x^2-x+1 = Ax^3 + Ax^2 + Ax + Bx^2 + Bx + B + Cx^3 + Cx + Dx^2 + D$
$x^2-x+1 = (A+C)x^3 + (A+B+D)x^2 + (A+B+C)x + (B+D)$
Comparing coefficients on both sides:
$1) A+C = 0$
$2) A+B+D = 1$
$3) A+B+C = -1$
$4) B+D = 1$
From $(1)$,$C = -A$. Substituting into $(3)$: $A+B-A = -1 \Rightarrow B = -1$.
Using $B = -1$ in $(4)$: $-1+D = 1 \Rightarrow D = 2$.
Using $B = -1$ and $D = 2$ in $(2)$: $A-1+2 = 1 \Rightarrow A = 0$.
Then $C = -A = 0$.
Thus,$A=0, B=-1, C=0, D=2$.
Calculating $A+2B+C+2D = 0 + 2(-1) + 0 + 2(2) = -2 + 4 = 2$.

(D) Given the partial fraction decomposition: $\frac{x-2}{x^2(2x-3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x-3}$.
Multiplying both sides by $x^2(2x-3)$,we get: $x-2 = Ax(2x-3) + B(2x-3) + Cx^2$.
Setting $x = 0$: $-2 = B(-3) \Rightarrow B = \frac{2}{3}$.
Setting $x = \frac{3}{2}$: $\frac{3}{2} - 2 = C(\frac{3}{2})^2$ $\Rightarrow -\frac{1}{2} = C(\frac{9}{4})$ $\Rightarrow C = -\frac{2}{9}$.
Comparing the coefficients of $x^2$: $0 = 2A + C$ $\Rightarrow 2A = -C = \frac{2}{9}$ $\Rightarrow A = \frac{1}{9}$.
Now,calculate $2(A-C) = 2(\frac{1}{9} - (-\frac{2}{9})) = 2(\frac{1+2}{9}) = 2(\frac{3}{9}) = 2(\frac{1}{3}) = \frac{2}{3}$.
Since $B = \frac{2}{3}$,we have $2(A-C) = B$.

(B) Given,$\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$
Equating the numerators:
$9x-7 = A(x^2+1) + (Bx+C)(x+3)$
$9x-7 = Ax^2 + A + Bx^2 + 3Bx + Cx + 3C$
$9x-7 = (A+B)x^2 + (3B+C)x + (A+3C)$
Comparing coefficients on both sides:
$A+B = 0$
$3B+C = 9$
$A+3C = -7$
From $A+B=0$,we get $B = -A$.
Substituting $B = -A$ into $3B+C=9$,we get $-3A+C=9$,so $C = 9+3A$.
Substituting $C = 9+3A$ into $A+3C=-7$:
$A + 3(9+3A) = -7$
$A + 27 + 9A = -7$
$10A = -34 \implies A = -\frac{34}{10} = -\frac{17}{5}$
Then $B = -A = \frac{17}{5}$
And $C = 9 + 3(-\frac{17}{5}) = 9 - \frac{51}{5} = \frac{45-51}{5} = -\frac{6}{5}$
Finally,$A+B+C = -\frac{17}{5} + \frac{17}{5} - \frac{6}{5} = -\frac{6}{5}$

(D) Let $x-3 = y$,so $x = y+3$.
Substituting this into the expression:
$\frac{(y+3)^3+3}{y^3} = \frac{y^3+9y^2+27y+27+3}{y^3} = \frac{y^3+9y^2+27y+30}{y^3} = 1 + \frac{9}{y} + \frac{27}{y^2} + \frac{30}{y^3}$.
Comparing this with $a + \frac{b}{y} + \frac{c}{y^2} + \frac{d}{y^3}$,we get:
$a = 1, b = 9, c = 27, d = 30$.
Now,calculate $(a+d)-(b+c)$:
$(1+30) - (9+27) = 31 - 36 = -5$.

(D) Given the partial fraction decomposition: $\frac{x^2-3}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1}$.
Multiplying both sides by $(x+2)(x^2+1)$,we get: $x^2-3 = A(x^2+1) + (Bx+C)(x+2)$.
To find $A$,set $x = -2$: $(-2)^2 - 3 = A((-2)^2 + 1) \implies 4-3 = A(4+1) \implies 1 = 5A \implies A = \frac{1}{5}$.
Expanding the right side: $x^2-3 = A x^2 + A + B x^2 + 2Bx + Cx + 2C = (A+B)x^2 + (2B+C)x + (A+2C)$.
Comparing coefficients of $x^2$: $A+B = 1 \implies \frac{1}{5} + B = 1 \implies B = \frac{4}{5}$.
Comparing coefficients of $x$: $2B+C = 0 \implies 2(\frac{4}{5}) + C = 0 \implies C = -\frac{8}{5}$.
Now calculate $3A+2B-C = 3(\frac{1}{5}) + 2(\frac{4}{5}) - (-\frac{8}{5}) = \frac{3}{5} + \frac{8}{5} + \frac{8}{5} = \frac{19}{5}$.

(C) Given the partial fraction decomposition: $\frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2}$.
Multiplying both sides by $(x+1)(x^2+2)$,we get: $x+3 = a(x^2+2) + (bx+c)(x+1)$.
Expanding the right side: $x+3 = ax^2 + 2a + bx^2 + bx + cx + c$.
Grouping terms by powers of $x$: $x+3 = (a+b)x^2 + (b+c)x + (2a+c)$.
Comparing coefficients on both sides:
$1$) $a+b = 0 \implies b = -a$
$2$) $b+c = 1$
$3$) $2a+c = 3$
Substitute $b = -a$ into $(2)$: $-a+c = 1 \implies c = a+1$.
Substitute $c = a+1$ into $(3)$: $2a + (a+1) = 3 \implies 3a = 2 \implies a = \frac{2}{3}$.
Then $b = -\frac{2}{3}$ and $c = \frac{2}{3} + 1 = \frac{5}{3}$.
Finally,calculate $a-b+c = \frac{2}{3} - (-\frac{2}{3}) + \frac{5}{3} = \frac{2}{3} + \frac{2}{3} + \frac{5}{3} = \frac{9}{3} = 3$.

(B) Let $y = x^2$. The expression becomes $\frac{y+1}{(y+2)(y+3)} = \frac{Ay+B}{y+2} + \frac{Cy+D}{y+3}$.
Using partial fractions for $\frac{y+1}{(y+2)(y+3)}$,we have $\frac{y+1}{(y+2)(y+3)} = \frac{P}{y+2} + \frac{Q}{y+3}$.
$y+1 = P(y+3) + Q(y+2)$.
For $y = -2$,$-2+1 = P(-2+3) \implies P = -1$.
For $y = -3$,$-3+1 = Q(-3+2) \implies -2 = -Q \implies Q = 2$.
So,$\frac{y+1}{(y+2)(y+3)} = \frac{-1}{y+2} + \frac{2}{y+3}$.
Substituting back $y = x^2$,we get $\frac{-1}{x^2+2} + \frac{2}{x^2+3}$.
Comparing this with $\frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$,we get $A=0, B=-1, C=0, D=2$.
Therefore,$A+B+C+D = 0 + (-1) + 0 + 2 = 1$.

(B) Given the partial fraction decomposition: $\frac{x+1}{x^3(x-1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{d}{x-1}$.
Multiplying both sides by $x^3(x-1)$,we get: $x+1 = ax^2(x-1) + bx(x-1) + c(x-1) + dx^3$.
Expanding the right side: $x+1 = a(x^3 - x^2) + b(x^2 - x) + c(x-1) + dx^3$.
Grouping the powers of $x$: $x+1 = (a+d)x^3 + (b-a)x^2 + (c-b)x - c$.
Comparing coefficients:
Constant term: $-c = 1 \implies c = -1$.
Coefficient of $x$: $c - b = 1 \implies -1 - b = 1 \implies b = -2$.
Coefficient of $x^2$: $b - a = 0 \implies a = b = -2$.
Coefficient of $x^3$: $a + d = 0 \implies d = -a = 2$.
Checking the values: $a = -2, b = -2, c = -1, d = 2$.
We observe that $a = b = 2c = -d$ since $-2 = -2 = 2(-1) = -(2)$.

(D) Given the partial fraction decomposition: $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$.
Multiplying both sides by $(x-1)^2(x^2+1)$,we get: $3x+1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2$.
Setting $x=1$: $3(1)+1 = B(1^2+1) \implies 4 = 2B \implies B=2$.
Expanding the right side: $3x+1 = A(x^3-x^2+x-1) + 2(x^2+1) + (Cx+D)(x^2-2x+1)$.
Comparing coefficients of $x^3$: $0 = A+C \implies C = -A$.
Comparing coefficients of $x^2$: $0 = -A+2+D-2C = -A+2+D+2A = A+D+2 \implies D = -A-2$.
Comparing constant terms: $1 = -A+2+D \implies -A+D = -1$.
Substituting $D = -A-2$: $-A-A-2 = -1 \implies -2A = 1 \implies A = -1/2$.
Then $C = 1/2$ and $D = -(-1/2)-2 = 1/2-2 = -3/2$.
We need to calculate $2(A-C+B+D) = 2(-1/2 - 1/2 + 2 - 3/2) = 2(-1 + 2 - 1.5) = 2(-0.5) = -1$.

(C) Given,$\frac{2x^3+x^2-5}{x^4-25}=\frac{Ax+B}{x^2-5}+\frac{Cx+1}{x^2+5}$
Since $x^4-25 = (x^2-5)(x^2+5)$,we have:
$2x^3+x^2-5 = (Ax+B)(x^2+5) + (Cx+1)(x^2-5)$
$2x^3+x^2-5 = Ax^3 + 5Ax + Bx^2 + 5B + Cx^3 - 5Cx + x^2 - 5$
$2x^3+x^2-5 = x^3(A+C) + x^2(B+1) + x(5A-5C) + (5B-5)$
Equating the coefficients of $x^3, x^2, x$ and the constant term:
$1) A+C = 2$
$2) B+1 = 1 \Rightarrow B = 0$
$3) 5A-5C = 0 \Rightarrow A = C$
$4) 5B-5 = -5 \Rightarrow 5(0)-5 = -5$ (Consistent)
Substituting $A=C$ into $A+C=2$,we get $2C=2$,so $C=1$ and $A=1$.
Thus,$(A, B, C) = (1, 0, 1)$.

(B) We can use partial fractions to decompose the expression: $\frac{2}{(1-x)(2-x)} = \frac{A}{1-x} + \frac{B}{2-x}$.
Solving for $A$ and $B$: $2 = A(2-x) + B(1-x)$.
For $x=1$,$2 = A(1) \implies A=2$.
For $x=2$,$2 = B(-1) \implies B=-2$.
So,$\frac{2}{(1-x)(2-x)} = \frac{2}{1-x} - \frac{2}{2-x} = 2(1-x)^{-1} - (1-\frac{x}{2})^{-1}$.
Expanding using the binomial series $(1-z)^{-1} = 1 + z + z^2 + z^3 + \dots$:
$2(1 + x + x^2 + x^3 + \dots) - (1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \dots)$.
The coefficient of $x^3$ is $2(1) - \frac{1}{8} = 2 - \frac{1}{8} = \frac{16-1}{8} = \frac{15}{8}$.