(a)

\begin{align} {3x + 1 \over (x - 2)(x + 5)} & = {A \over x - 2} + {B \over x + 5} \\ & = {A(x + 5) \over (x - 2)(x + 5)} + {B(x - 2) \over (x - 2)(x + 5)} \\ & = {A(x + 5) + B(x - 2) \over (x - 2)(x + 5} \\ \\ \text{Comparing } & \text{the numerator,} \\ 3x + 1 & = A(x + 5) + B(x - 2) \\ \\ \text{Let } & x = -5, \\ 3(-5) + 1 & = A(-5 + 5) + B(-5 - 2) \\ -14 & = A(0) + B(-7) \\ -14 & = -7B \\ {-14 \over -7} & = B \\ 2 & = B \\ \\ \text{Let } & x = 2, \\ 3(2) + 1 & = A(2 + 5) + 2(2 - 2) \\ 7 & = A(7) + 2(0) \\ 7 & = 7A \\ {7 \over 7} & = A \\ 1 & = A \\ \\ \therefore {3x + 1 \over (x - 2)(x + 5)} & = {1 \over x - 2} + {2 \over x + 5} \end{align}

(b)

\begin{align} {x - 2 \over (x + 2)(x + 1)} & = {A \over x + 2} + {B \over x + 1} \\ & = {A(x + 1) \over (x + 2)(x + 1)} + {B(x + 2) \over (x + 2)(x + 1)} \\ & = {A(x + 1) + B(x + 2) \over (x + 2)(x + 1)} \\ \\ \text{Comparing the} & \text{ numerator,} \\ x - 2 & = A(x + 1) + B(x + 2) \\ \\ \text{Let } & x = -1, \\ (-1) - 2 & = A(-1 + 1) + B(-1 + 2) \\ -3 & = A(0) + B(1) \\ -3 & = B \\ \\ \text{Let } & x = - 2, \\ (-2) - 2 & = A(-2 + 1) + (-3)(-2 + 2) \\ -4 & = A(-1) + (-3)(0) \\ -4 & = -A \\ 4 & = A \\ \\ \therefore {x - 2 \over (x + 2)(x + 1)} & = {4 \over x + 2} + {-3 \over x + 1} \\ & = {4 \over x + 2} - {3 \over x + 1} \end{align}

(c)

\begin{align} {x + 5 \over x^2 + x} & = {x + 5 \over x(x + 1)} \\ & = {A \over x} + {B \over x + 1} \\ & = {A(x + 1) \over x(x + 1)} + {B(x) \over x(x + 1)} \\ & = {A(x + 1) + B(x) \over x(x + 1)} \\ \\ \text{Comparing } & \text{the numerator,} \\ x + 5 & = A(x + 1) + B(x) \\ \\ \text{Let } & x = -1, \\ (-1) + 5 & = A(-1 + 1) + B(-1) \\ 4 & = A(0) - B \\ 4 & = -B \\ -4 & = B \\ \\ \text{Let } & x = 0, \\ (0) + 5 & = A(0 + 1) + (-4)(0) \\ 5 & = A + 0 \\ 5 & = A \\ \\ \therefore {x + 5 \over x^2 + x} & = {5 \over x} + {-4 \over x + 1} \\ & = {5 \over x} - {4 \over x + 1} \end{align}

(d)

\begin{align} {x + 10 \over x^2 - 4} & = {x + 10 \over (x + 2)(x - 2)} \\ & = {A \over x + 2} + {B \over x - 2} \\ & = {A(x - 2) \over (x + 2)(x - 2)} + {B(x + 2) \over (x + 2)(x - 2)} \\ & = {A(x - 2) + B(x + 2) \over (x + 2)(x - 2)} \\ \\ \text{Comparing } & \text{the numerator,} \\ x + 10 & = A(x - 2) + B(x + 2) \\ \\ \text{Let } & x = 2, \\ \\ (2) + 10 & = A(2 - 2) + B(2 + 2) \\ 12 & = A(0) + B(4) \\ 12 & = 4B \\ {12 \over 4} & = B \\ 3 & = B \\ \\ \text{Let } & x = -2, \\ (-2) + 10 & = A(-2 - 2) + (3)(-2 + 2) \\ 8 & = A(-4) + (3)(0) \\ 8 & = -4A \\ {8 \over -4} & = A \\ -2 & = A \\ \\ \therefore {x + 5 \over x^2 + x} & = {-2 \over x + 2} + {3 \over x - 2} \\ & = -{2 \over x + 2} + {3 \over x - 2} \\ & = {3 \over x - 2} - {2 \over x + 2} \end{align}

(a)

\begin{align} {x + 2 \over (x + 1)^2} & = {A \over x + 1} + {B \over (x + 1)^2} \\ & = {A(x + 1) \over (x + 1)^2} + {B \over (x + 1)^2} \\ & = {A(x + 1) + B \over (x + 1)^2} \\ \\ \text{Comparing } & \text{the numerator,} \\ x + 2 & = A(x + 1) + B \\ \\ \text{Let } & x = - 1, \\ (-1) + 2 & = A(-1 + 1) + B \\ 1 & = A(0) + B \\ 1 & = B \\ \\ \text{Let } & x = 0, \\ (0) + 2 & = A(0 + 1) + (1) \\ 2 & = A(1) + 1 \\ 2 & = A + 1 \\ 2- 1 & = A \\ 1 & = A \\ \\ \therefore {x + 2 \over (x + 1)^2} & = {1 \over x + 1} + {1 \over (x + 1)^2} \end{align}

(b)

\begin{align} {x^2 - 8x + 44 \over (x + 2)(x - 2)^2} & = {A \over x + 2} + {B \over x - 2} + {C \over (x - 2)^2} \\ & = {A(x - 2)^2 \over (x + 2)(x - 2)^2} + {B(x + 2)(x - 2) \over (x + 2)(x - 2)^2} + {C(x + 2) \over (x + 2)(x - 2)^2} \\ & = {A(x - 2)^2 + B(x + 2)(x - 2) + C(x + 2) \over (x + 2)(x - 2)^2} \\ \\ \text{Comparing } & \text{the numerator,} \\ x^2 - 8x + 44 & = A(x - 2)^2 + B(x + 2)(x - 2) + C(x + 2) \\ \\ \text{Let } & x = 2, \\ (2)^2 - 8(2) + 44 & = A(2 - 2)^2 + B(2 + 2)(2 - 2) + C(2 + 2) \\ 32 & = A(0)^2 + B(4)(0) + C(4) \\ 32 & = 4C \\ {32 \over 4} & = C \\ 8 & = C \\ \\ \text{Let } & x = -2, \\ (-2)^2 - 8(-2) + 44 & = A(-2 - 2)^2 + B(-2 + 2)(-2 -2) + 8(-2 + 2) \\ 64 & = A(-4)^2 + B(0)(-4) + 8(0) \\ 64 & = 16A \\ {64 \over 16} & = A \\ 4 & = A \\ \\ \text{Let } & x = 0, \\ (0)^2 - 8(0) + 44 & = 4(0 - 2)^2 + B(0 + 2)(0 - 2) + 8(0 + 2) \\ 44 & = 4(-2)^2 + B(2)(-2) + 8(2) \\ 44 & = 16 - 4B + 16 \\ 4B & = 16 + 16 - 44 \\ 4B & = -12 \\ B & = {-12 \over 4} \\ & = -3 \\ \\ \therefore {x^2 - 8x + 44 \over (x + 2)(x - 2)^2} & = {4 \over x + 2} + {-3 \over x - 2} + {8 \over (x - 2)^2} \\ & = {4 \over x + 2} - {3 \over x - 2} + {8 \over (x - 2)^2} \end{align}

(c)

\begin{align} {14 + 7x - 3x^2 \over x^2(x + 2)} & = {A \over x} + {B \over x^2} + {C \over x + 2} \\ & = {A(x)(x + 2) \over x^2(x + 2)} + {B(x + 2) \over x^2(x + 2)} + {C(x^2) \over x^2(x + 2)} \\ & = {A(x)(x + 2) + B(x + 2) + C(x^2) \over x^2(x + 2)} \\ \\ \text{Comparing } & \text{the numerator,} \\ 14 + 7x - 3x^2 & = A(x)(x + 2) + B(x + 2) + C(x^2) \\ \\ \text{Let } & x = 0, \\ 14 + 7(0) - 3(0)^2 & = A(0)(0 + 2) + B(0 + 2) + C(0)^2 \\ 14 & = B(2) \\ 14 & = 2B \\ {14 \over 2} & = B \\ 7 & = B \\ \\ \text{Let } & x = -2, \\ 14 + 7(-2) - 3(-2)^2 & = A(-2)(-2 + 2) + 7(-2 + 2) + C(-2)^2 \\ -12 & = A(-2)(0) + 7(0) + C(4) \\ -12 & = 4C \\ {-12 \over 4} & = C \\ -3 & = C \\ \\ \text{Let } & x = 1, \\ 14 + 7(1) - 3(1)^2 & = A(1)(1 + 2) + 7(1 + 2) - 3(1)^2 \\ 18 & = A(1)(3) + 7(3) - 3(1) \\ 18 & = 3A + 21 - 3 \\ 18 & = 3A + 18 \\ 18 - 18 & = 3A \\ 0 & = 3A \\ {0 \over 3} & = A \\ 0 & = A \\ \\ \therefore {14 + 7x - 3x^2 \over x^2(x + 2)} & = {0 \over x} + {7 \over x^2} + {-3 \over x + 2} \\ & = {7 \over x^2} - {3 \over x + 2} \end{align}

(d)

\begin{align} {4x^2 - 9x + 7 \over x(x - 1)^2} & = {A \over x} + {B \over x - 1} + {C \over (x - 1)^2} \\ & = {A(x - 1)^2 \over x(x - 1)^2} + {B(x)(x - 1) \over x(x - 1)^2} + {C(x) \over x(x - 1)^2} \\ & = {A(x - 1)^2 + B(x)(x - 1) + C(x) \over x(x - 1)^2} \\ \\ \text{Comparing } & \text{the numerator,} \\ 4x^2 - 9x + 7 & = A(x - 1)^2 + B(x)(x - 1) + C(x) \\ \\ \text{Let } & x = 1, \\ 4(1)^2 - 9(1) + 7 & = A(1 - 1)^2 + B(1)(1 - 1) + C(1) \\ 2 & = A(0) + B(1)(0) + C \\ 2 & = C \\ \\ \text{Let } & x = 0, \\ 4(0)^2 - 9(0) + 7 & = A(0 - 1)^2 + B(0)(0 - 1) + 2(0) \\ 7 & = A(-1)^2 + B(0)(-1) \\ 7 & = A(1) \\ 7 & = A \\ \\ \text{Let } & x = 2, \\ 4(2)^2 - 9(2) + 7 & = 7(2 - 1)^2 + B(2)(2 - 1) + 2(2) \\ 5 & = 7(1)^2 + B(2)(1) + 4 \\ 5 & = 7 + 2B + 4 \\ 5 - 7 - 4 & = 2B \\ -6 & = 2B \\ {-6 \over 2} & = B \\ -3 & = B \\ \\ \therefore {4x^2 - 9x + 7 \over x(x - 1)^2} & = {7 \over x} + {-3 \over x - 1} + {2 \over (x - 1)^2} \\ & = {7 \over x} - {3 \over x - 1} + {2 \over (x - 1)^2} \end{align}

(a)

\begin{align} {8x^2 - 11x + 5 \over (2x - 3)(x^2 + 1)} & = {A \over 2x - 3} + {Bx + C \over x^2 + 1} \\ & = {A(x^2 + 1) \over (2x - 3)(x^2 + 1)} + {(Bx + C)(2x - 3) \over (2x - 3)(x^2 + 1)} \\ & = {A(x^2 + 1) + (Bx + C)(2x - 3) \over (2x - 3)(x^2 + 1)} \\ \\ \text{Comparing } & \text{the numerator,} \\ 8x^2 - 11x + 5 & = A(x^2 + 1) + (Bx + C)(2x - 3) \\ \\ \text{Let } & x = 1.5, \\ 8(1.5)^2 - 11(1.5) + 5 & = A[(1.5)^2 + 1] + [B(1.5) + C][2(1.5) - 3] \\ 6.5 & = A(3.25) + (1.5B + C)(3 - 3) \\ 6.5 & = 3.25A + (1.5B + C)(0) \\ 6.5 & = 3.25A \\ {6.5 \over 3.25} & = A \\ 2 & = A \\ \\ \text{Let } & x = 0, \\ 8(0)^2 - 11(0) + 5 & = 2[(0)^2 + 1] + [B(0) + C][2(0) - 3] \\ 5 & = 2(1) + (C)(-3) \\ 5 & = 2 - 3C \\ 5 - 2 & = -3C \\ 3 & = -3C \\ {3 \over -3} & = C \\ -1 & = C \\ \\ \text{Let } & x = 1, \\ 8(1)^2 - 11(1) + 5 & = 2[(1)^2 + 1] + [B(1) - 1][2(1) - 3] \\ 2 & = 2(2) + (B - 1)(-1) \\ 2 & = 4 - B + 1 \\ B & = 4 + 1 - 2 \\ & = 3 \\ \\ {8x^2 - 11x + 5 \over (2x - 3)(x^2 + 1)} & = {2 \over 2x - 3} + {(3)x + (-1) \over x^2 + 1} \\ & = {2 \over 2x - 3} + {3x - 1 \over x^2 + 1} \end{align}

(b)

\begin{align} {x^2 + 2x + 15 \over x(x^2 + 3)} & = {A \over x} + {Bx + C \over x^2 + 3} \\ & = {A(x^2 + 3) \over x(x^2 + 3)} + {(Bx + C)(x) \over x(x^2 + 3)} \\ & = {A(x^2 + 3) + (Bx + C)(x) \over x(x^2 + 3)} \\ \\ \text{Comparing } & \text{the numerator,} \\ x^2 + 2x + 15 & = A(x^2 + 3) + (Bx + C)(x) \\ \\ \text{Let } & x = 0, \\ (0)^2 + 2(0) + 15 & = A[(0)^2 + 3] + [B(0) + C](0) \\ 15 & = A(3) + (C)(0) \\ 15 & = 3A \\ {15 \over 3} & = A \\ 5 & = A \\ \\ x^2 + 2x + 15 & = 5(x^2 + 3) + (Bx + C)(x) \\ x^2 + 2x + 15 & = 5x^2 + 15 + Bx^2 + Cx \\ x^2 + 2x + 15 & = 5x^2 + Bx^2 + Cx + 15 \\ x^2 + 2x + 15 & = (5 + B)x^2 + Cx + 15 \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ 1 & = 5 + B \\ 1 - 5 & = B \\ -4 & = B \\ \\ \text{Comparing } & \text{coefficients of } x, \\ 2 & = C \\ \\ \therefore {x^2 + 2x + 15 \over x(x^2 + 3)} & = {A \over x} + {Bx + C \over x^2 + 3} \\ & = {5 \over x} + {(-4)x + (2) \over x^2 + 3} \\ & = {5 \over x} + {2 - 4x \over x^2 + 3} \end{align}

(c)

\begin{align} {7x^2 - 9x + 29 \over (x - 3)(x^2 + 4)} & = {A \over x - 3} + {Bx + C \over x^2 + 4} \\ & = {A(x^2 + 4) \over (x - 3)(x^2 + 4)} + {(Bx + C)(x - 3) \over (x - 3)(x^2 + 4)} \\ & = {A(x^2 + 4) + (Bx + C)(x - 3) \over (x - 3)(x^2 + 4)} \\ \\ \text{Comparing } & \text{the numerator,} \\ 7x^2 - 9x + 29 & = A(x^2 + 4) + (Bx + C)(x - 3) \\ \\ \text{Let } & x = 3, \\ 7(3)^2 - 9(3) + 29 & = A[(3)^2 + 4] + [B(3) + C](3 - 3) \\ 65 & = A(13) + (3B + C)(0) \\ 65 & = 13A \\ {65 \over 13} & = A \\ 5 & = A \\ \\ \text{Let } & x = 0, \\ 7(0)^2 - 9(0) + 29 & = 5[(0)^2 + 4] + [B(0) + C](0 - 3) \\ 29 & = 5(4) + (C)(-3) \\ 29 & = 20 - 3C \\ 29 - 20 & = -3C \\ 9 & = -3C \\ {9 \over -3} & = C \\ -3 & = C \\ \\ \text{Let } & x = 1, \\ 7(1)^2 - 9(1) + 29 & = 5[(1)^2 + 4] + [B(1) + (-3)](1 - 3) \\ 27 & = 5(5) + (B - 3)(-2) \\ 27 & = 25 - 2B + 6 \\ 27 - 25 - 6 & = -2B \\ -4 & = -2B \\ {-4 \over -2} & = B \\ 2 & = B \\ \\ {7x^2 - 9x + 29 \over (x - 3)(x^2 + 4)} & = {A \over x - 3} + {Bx + C \over x^2 + 4} \\ & = {5 \over x - 3} + {(2)x + (-3) \over x^2 + 4} \\ & = {5 \over x - 3} + {2x - 3 \over x^2 + 4} \end{align}

(d)

\begin{align} {x^2 + 5x + 6 \over (x - 1)(x^2 + 5)} & = {A \over x - 1} + {Bx + C \over x^2 + 5} \\ & = {A(x^2 + 5) \over (x - 1)(x^2 + 5)} + {(Bx + C)(x - 1) \over (x - 1)(x^2 + 5)} \\ & = {A(x^2 + 5) + (Bx + C)(x - 1) \over (x - 1)(x^2 + 5)} \\ \\ \text{Comparing } & \text{the numerator,} \\ x^2 + 5x + 6 & = A(x^2 + 5) + (Bx + C)(x - 1) \\ \\ \text{Let } & x = 1, \\ (1)^2 + 5(1) + 6 & = A[(1)^2 + 5] + [B(1) + C](1 - 1) \\ 12 & = A(6) + (B + C)(0) \\ 12 & = 6A \\ {12 \over 6} & = A \\ 2 & = A \\ \\ \text{Let } & x = 0, \\ (0)^2 + 5(0) + 6 & = 2[(0)^2 + 5] + [B(0) + C](0 - 1) \\ 6 & = 2(5) + (C)(-1) \\ 6 & = 10 - C \\ C & = 10 - 6 \\ & = 4 \\ \\ \text{Let } & x = 2, \\ (2)^2 + 5(2) + 6 & = 2[(2)^2 + 5] + [B(2) + (4)](2 - 1) \\ 20 & = 2(9) + (2B + 4)(1) \\ 20 & = 18 + 2B + 4 \\ 20 - 18 - 4 & = 2B \\ -2 & = 2B \\ {-2 \over 2} & = B \\ -1 & = B \\ \\ {x^2 + 5x + 6 \over (x - 1)(x^2 + 5)} & = {A \over x - 1} + {Bx + C \over x^2 + 5} \\ & = {2 \over x - 1} + {(-1)x + 4 \over x^2 + 5} \\ & = {2 \over x - 1} + {4 - x \over x^2 + 5} \end{align}

(a)

The degree of the numerator is 1 while the degree of the denominator is 2. Thus, it is a proper fraction.

(b)

The degree of the numerator is 1 while the degree of the denominator is 2. Thus, it is a proper fraction.

(c)

The degree of the numerator is 2 while the degree of the denominator is 2. Thus, it is an improper fraction. $$ \require{enclose} \begin{array}{rll} 4 \phantom{0.}\\ x^2 - 2 \enclose{longdiv}{ 4x^2 + 3 \phantom{0}}\kern-.2ex \\ -\underline{( 4x^2 - 8 ){\phantom{.}}} \\ 11 \phantom{0.} \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ \therefore {4x^2 + 3 \over x^2 - 2} & = 4 + {11 \over x^2 - 2} \end{align}

(d)

\begin{align} {(x + 1)(x^3 - 2) \over x^2(2x - 3)} & = {x^4 - 2x + x^3 - 2 \over 2x^3 - 3x^2} \\ & = {x^4 + x^3 - 2x - 2 \over 2x^3 - 3x^2} \end{align}
The degree of the numerator is 4 while the degree of the denominator is 3. Thus, it is an improper fraction.
$$ \require{enclose} \begin{array}{rll} {1 \over 2}x + {5 \over 4} \phantom{00000000000000.}\\ 2x^3 - 3x^2 \enclose{longdiv}{ x^4 + x^3 + 0x^2 - 2x - 2 \phantom{0}}\kern-.2ex \\ -\underline{( x^4 - {3 \over 2}x^3 ){\phantom{00000000000.}}} \\ {5 \over 2}x^3 + 0x^2 - 2x - 2 \phantom{0.} \\ -\underline{ ( {5 \over 2}x^3 - {15 \over 4}x^2 )\phantom{0000000} } \\ {15 \over 4}x^2 - 2x - 2 \phantom{0.} \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ \therefore {(x + 1)(x^3 - 2) \over x^2(2x - 3)} & = {1 \over 2}x + {5 \over 4} + {{15 \over 4}x^2 - 2x - 2 \over x^2(2x - 3)} \end{align}

(a)

\begin{align} {x + 3 \over (x - 2)(x - 4)} & = {A \over x - 2} + {B \over x - 4} \\ & = {A(x - 4) \over (x - 2)(x - 4)} + {B(x - 2) \over (x - 2)(x - 4)} \\ & = {A(x - 4) + B(x - 2) \over (x - 2)(x - 4)} \\ \\ \text{Comparing } & \text{the numerator,} \\ x + 3 & = A(x - 4) + B(x - 2) \\ \\ \text{Let } & x = 4, \\ (4) + 3 & = A(4 - 4) + B(4 - 2) \\ 7 & = A(0) + B(2) \\ 7 & = 2B \\ {7 \over 2} & = B \\ \\ \text{Let } & x = 2, \\ (2) + 3 & = A(2 - 4) + {7 \over 2}(2 - 2) \\ 5 & = A(-2) + {7 \over 2}(0) \\ 5 & = -2A \\ -{5 \over 2} & = A \\ \\ \therefore {x + 3 \over (x - 2)(x - 4)} & = {-{5 \over 2} \over x - 2} + {{7 \over 2} \over x - 4} \\ & = -{5 \over 2(x - 2)} + {7 \over 2(x - 4)} \\ & = {7 \over 2(x - 4)} -{5 \over 2(x - 2)} \end{align}

(b)

\begin{align} {x + 3 \over (x - 2)(x^2 - 4)} & = {x + 3 \over (x - 2)(x - 2)(x + 2)} \\ & = {x + 3 \over (x - 2)^2 (x + 2)} \\ & = {A \over x - 2} + {B \over (x - 2)^2} + {C \over x + 2} \\ & = {A(x - 2)(x + 2) \over (x - 2)^2 (x + 2)} + {B(x + 2) \over (x - 2)^2 (x + 2)} + {C(x - 2)^2 \over (x - 2)^2 (x + 2)} \\ & = {A(x - 2)(x + 2) + B(x + 2) + C(x - 2)^2 \over (x - 2)^2 (x + 2)} \\ \\ \\ \text{Comparing } & \text{the numerator,} \\ x + 3 & = A(x - 2)(x + 2) + B(x + 2) + C(x - 2)^2 \\ \\ \text{Let } & x = 2, \\ 2 + 3 & = A(2 - 2)(2 + 2) + B(2 + 2) + C(2 - 2)^2 \\ 5 & = A(0)(4) + B(4) + C(0)^2 \\ 5 & = 4B \\ {5 \over 4} & = B \\ \\ \text{Let } & x = -2, \\ (-2) + 3 & = A(-2 - 2)(-2 + 2) + {5 \over 4}(-2 + 2) + C(-2 - 2)^2 \\ 1 & = A(-4)(0) + {5 \over 4}(0) + C(-4)^2 \\ 1 & = C(16) \\ 1 & = 16C \\ {1 \over 16} & = C \\ \\ \text{Let } & x = 0 , \\ 0 + 3 & = A(0 - 2)(0 + 2) + {5 \over 4}(0 + 2) + {1 \over 16}(0 - 2)^2 \\ 3 & = A(-2)(2) + {5 \over 4}(2) + {1 \over 16}(4) \\ 3 & = -4A + {5 \over 2} + {1 \over 4} \\ 4A & = {5 \over 2} + {1 \over 4} - 3 \\ 4A & = -{1 \over 4} \\ A & = {-{1 \over 4} \over 4} \\ & = -{1 \over 16} \\ \\ {x + 3 \over (x - 2)(x^2 - 4)} & = {-{1 \over 16} \over x - 2} + {{5 \over 4} \over (x - 2)^2} + {{1 \over 16} \over x + 2} \\ & = -{1 \over 16(x - 2)} + {5 \over 4(x - 2)^2} + {1 \over 16(x + 2)} \\ & = {1 \over 16(x + 2)} -{1 \over 16(x - 2)} + {5 \over 4(x - 2)^2} \end{align}

(c)

\begin{align} {x + 3 \over (x - 2)(x^2 + 4)} & = {A \over x - 2} + {Bx + C \over x^2 + 4} \\ & = {A(x^2 + 4) \over (x - 2)(x^2 + 4)} + {(Bx + C)(x - 2) \over (x - 2)(x^2 + 4)} \\ & = {A(x^2 + 4) + (Bx + C)(x - 2) \over (x - 2)(x^2 + 4)} \\ \\ \text{Comparing } & \text{the numerator,} \\ x + 3 & = A(x^2 + 4) + (Bx + C)(x - 2) \\ \\ \text{Let } & x = 2, \\ 2 + 3 & = A[(2)^2 + 4] + [B(2) + C](2 - 2) \\ 5 & = A(8) + (2B + C)(0) \\ 5 & = 8A \\ {5 \over 8} & = A \\ \\ \text{Let } & x = 0, \\ 0 + 3 & = {5 \over 8}(0^2 + 4) + [B(0) + C](0 - 2) \\ 3 & = {5 \over 8}(4) + (C)(-2) \\ 3 & = {5 \over 2} - 2C \\ 2C & = {5 \over 2} - 3 \\ & = -{1 \over 2} \\ C & = {-{1 \over 2} \over 2} \\ & = -{1 \over 4} \\ \\ \text{Let } & x = 1, \\ 1 + 3 & = {5 \over 8}(1^2 + 4) + \left[ B(1) + \left(-{1 \over 4}\right)\right](1 - 2) \\ 4 & = {5 \over 8} (5) + \left(B - {1 \over 4}\right)(-1) \\ 4 & = {25 \over 8} - B + {1 \over 4} \\ B & = {25 \over 8} + {1 \over 4} - 4 \\ & = -{5 \over 8} \\ \\ \therefore {x + 3 \over (x - 2)(x^2 + 4)} & = {{5 \over 8} \over x - 2} + {-{5 \over 8}x + (-{1 \over 4}) \over x^2 + 4} \\ & = {5 \over 8(x - 2)} + {-{5 \over 8}x -{1 \over 4} \over x^2 + 4} \\ & = {5 \over 8(x - 2)} + {-{5 \over 8}x -{2 \over 8} \over x^2 + 4} \\ & = {5 \over 8(x - 2)} + {(-{1 \over 8})5x + 2 \over x^2 + 4} \\ & = {5 \over 8(x - 2)} - {5x + 2 \over 8(x^2 + 4)} \end{align}

(a)

\begin{align} (x + 2)(x + 3) & = x^2 + 3x + 2x + 6 \\ & = x^2 + 5x + 6 \end{align} $$ \require{enclose} \begin{array}{rll} 2x + 5 \phantom{000000000000.}\\ x^2 + 5x + 6\enclose{longdiv}{ 2x^3 + 15x^2 + 39x + 33 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 + 10x^2 + 12x ){\phantom{0000.}}} \\ 5x^2 + 27x + 33 \phantom{0.} \\ -\underline{ ( 5x^2 + 25x + 30 )\phantom{0} } \\ 2x + 3 \phantom{0.} \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ {2x^3 + 15x^2 + 39x + 33 \over (x + 2)(x + 3)} & = 2x + 5 + {2x + 3 \over (x + 2)(x + 3)} \\ \\ {2x + 3 \over (x + 2)(x + 3)} & = {A \over x + 2} + {B \over x + 3} \\ & = {A(x + 3) \over (x + 2)(x + 3)} + {B(x + 2) \over (x + 2)(x + 3)} \\ & = {A(x + 3) + B(x + 2) \over (x + 2)(x + 3)} \\ \\ \text{Comparing } & \text{the numerator,} \\ 2x + 3 & = A(x + 3) + B(x + 2) \\ \\ \text{Let } & x = -3, \\ 2(-3) + 3 & = A(-3 + 3) + B(-3 + 2) \\ -3 & = A(0) + B(-1) \\ -3 & = -B \\ 3 & = B \\ \\ \text{Let } & x = -2, \\ 2(-2) + 3 & = A(-2 + 3) + 3(-2 + 2) \\ -1 & = A(1) + 3(0) \\ -1 & = A \\ \\ {2x + 3 \over (x + 2)(x + 3)} & = {-1 \over x + 2} + {3 \over x + 3} \\ & = -{1 \over x + 2} + {3 \over x + 3} \\ \\ \therefore {2x^3 + 15x^2 + 39x + 33 \over (x + 2)(x + 3)} & = 2x + 5 -{1 \over x + 2} + {3 \over x + 3} \end{align}

(b)

\begin{align} (x + 1)(x - 2) & = x^2 - 2x + x - 2 \\ & = x^2 - x - 2 \end{align} $$ \require{enclose} \begin{array}{rll} 4x \phantom{000000000000000.}\\ x^2 - x - 2\enclose{longdiv}{ 4x^3 - 4x^2 - 16x + 7 \phantom{0}}\kern-.2ex \\ -\underline{( 4x^3 - 4x^2 - 8x ){\phantom{0000.}}} \\ -8x + 7 \phantom{0.} \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ {4x^3 - 4x^2 - 16x + 7 \over (x + 1)(x - 2)} & = 4x + {-8x + 7 \over (x + 1)(x - 2)} \\ \\ {-8x + 7 \over (x + 1)(x - 2)} & = {A \over x + 1} + {B \over x - 2} \\ & = {A(x - 2) \over (x + 1)(x - 2)} + {B(x + 1) \over (x + 1)(x - 2)} \\ & = {A(x - 2) + B(x + 1) \over (x + 1)(x - 2)} \\ \\ \text{Comparing } & \text{the numerator,} \\ -8x + 7 & = A(x - 2) + B(x + 1) \\ \\ \text{Let } & x = 2, \\ -8(2) + 7 & = A(2 - 2) + B(2 + 1) \\ -9 & = A(0) + B(3) \\ -9 & = 3B \\ {-9 \over 3} & = B \\ -3 & = B \\ \\ \text{Let } & x = -1, \\ -8(-1) + 7 & = A(-1 - 2) + (-3)(-1 + 1) \\ 15 & = A(-3) + (-3)(0) \\ 15 & = -3A \\ {15 \over -3} & = A \\ -5 & = A \\ \\ {-8x + 7 \over (x + 1)(x - 2)} & = {-5 \over x + 1} + {-3 \over x - 2} \\ & = -{5 \over x + 1} - {3 \over x - 2} \\ \\ {4x^3 - 4x^2 - 16x + 7 \over (x + 1)(x - 2)} & = 4x -{5 \over x + 1} - {3 \over x - 2} \end{align}

(c)

\begin{align} (2x + 3)(x - 1)^2 & = (2x + 3)(x^2 - 2x + 1) \\ & = 2x^3 - 4x^2 + 2x + 3x^2 - 6x + 3 \\ & = 2x^3 - 4x^2 + 3x^2 + 2x - 6x + 3 \\ & = 2x^3 - x^2 - 4x + 3 \end{align} $$ \require{enclose} \begin{array}{rll} 3 \phantom{000000000000000.}\\ 2x^3 - x^2 - 4x + 3 \enclose{longdiv}{ 6x^3 - 5x^2 - 19x + 28 \phantom{0}}\kern-.2ex \\ -\underline{( 6x^3 - 3x^2 - 12x + 9 ){\phantom{0}}} \\ -2x^2 - 7x + 19 \phantom{0.} \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ {6x^3 - 5x^2 - 19x + 28 \over (2x + 3)(x - 1)^2} & = 3 + {-2x^2 - 7x + 19 \over (2x + 3)(x - 1)^2} \\ \\ {-2x^2 - 7x + 19 \over (2x + 3)(x - 1)^2} & = {A \over 2x + 3} + {B \over x - 1} + {C \over (x - 1)^2} \\ & = {A(x - 1)^2 \over (2x + 3)(x - 1)^2} + {B(2x + 3)(x - 1) \over (2x + 3)(x - 1)^2} + {C(2x + 3) \over (2x + 3)(x - 1)^2} \\ & = {A(x - 1)^2 + B(2x + 3)(x - 1) + C(2x + 3) \over (2x + 3)(x - 1)^2} \\ \\ \text{Comparing } & \text{the numerator,} \\ -2x^2 - 7x + 19 & = A(x - 1)^2 + B(2x + 3)(x - 1) + C(2x + 3) \\ \\ \text{Let } & x = 1, \\ -2(1)^2 - 7(1) + 19 & = A(1 - 1)^2 + B[2(1) + 3](1 - 1) + C[2(1) + 3] \\ 10 & = A(0)^2 + B(5)(0) + C(5) \\ 10 & = 5C \\ {10 \over 5} & = C \\ 2 & = C \\ \\ \text{Let } & x = -1.5, \\ -2(-1.5)^2 - 7(-1.5) + 19 & = A(-1.5 - 1)^2 + B[2(-1.5) + 3](-1.5 - 1) + 2[2(-1.5) + 3] \\ 25 & = A(6.25) + B(0)(-2.5) + 2(0) \\ 25 & = 6.25A \\ {25 \over 6.25} & = A \\ 4 & = A \\ \\ \text{Let } & x = 0, \\ 2(0)^2 - 7(0) + 19 & = 4(0 - 1)^2 + B[2(0) + 3](0 - 1) + 2[2(0) + 3] \\ 19 & = 4(-1)^2 + B(3)(-1) + 2(3) \\ 19 & = 4 - 3B + 6 \\ 3B & = 4 + 6 - 19 \\ 3B & = -9 \\ B & = {-9 \over 3} \\ & = -3 \\ \\ {-2x^2 - 7x + 19 \over (2x + 3)(x - 1)^2} & = {4 \over 2x + 3} + {-3 \over x - 1} + {2 \over (x - 1)^2} \\ & = {4 \over 2x + 3} - {3 \over x - 1} + {2 \over (x - 1)^2} \\ \\ {6x^3 - 5x^2 - 19x + 28 \over (2x + 3)(x - 1)^2} & = 3 + {4 \over 2x + 3} - {3 \over x - 1} + {2 \over (x - 1)^2} \end{align}

(d)

\begin{align} x(x^2 + 1) & = x^3 + x \end{align} $$ \require{enclose} \begin{array}{rll} 4x^2 + 2x - 1 \phantom{00000000000000.}\\ x^3 + 0x^2 + x \enclose{longdiv}{ 4x^5 + 2x^4 + 3x^3 - x^2 - x + 1 \phantom{0}}\kern-.2ex \\ -\underline{( 4x^5 + 0x^4 + 4x^3 ){\phantom{00000000000}}} \\ 2x^4 - x^3 - x^2 - x + 1 \phantom{00.} \\ -\underline{( 2x^4 + 0x^3 + 2x^2 ){\phantom{000000}}} \\ -x^3 - 3x^2 - x + 1 \phantom{00} \\ -\underline{( - x^3 + 0x^2 - x ){\phantom{0000.}}} \\ - 3x^2 \phantom{+9x} + 1 \phantom{00} \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ {4x^5 + 2x^4 + 3x^3 - x^2 -x + 1 \over x(x^2 + 1)} & = 4x^2 + 2x - 1 + {-3x^2 + 1 \over x(x^2 + 1)} \\ \\ {-3x^2 + 1 \over x(x^2 + 1)} & = {A \over x} + {Bx + C \over x^2 + 1} \\ & = {A(x^2 + 1) \over x(x^2 + 1)} + {(Bx + C)(x) \over x(x^2 + 1)} \\ & = {A(x^2 + 1) + (Bx + C)(x) \over x(x^2 + 1)} \\ \\ \text{Comparing } & \text{the numerator,} \\ -3x^2 + 1 & = A(x^2 + 1) + (Bx + C)(x) \\ \\ \text{Let } & x = 0, \\ -3(0)^2 + 1 & = A(0^2 + 1) + [B(0) + C](0) \\ 1 & = A(1) + (C)(0) \\ 1 & = A \\ \\ -3x^2 + 1 & = (1)(x^2 + 1) + (Bx + C)(x) \\ -3x^2 + 1 & = x^2 + 1 + Bx^2 + Cx \\ -3x^2 - x^2 + 1 - 1 & = Bx^2 + Cx \\ -4x^2 & = Bx^2 + Cx \\ -4x^2 + 0x & = Bx^2 + Cx \\ \\ \text{Comparing} & \text{ coefficients of } x^2, \\ -4 & = B \\ \\ \text{Comparing} & \text{ coefficients of } x, \\ 0 & = C \\ \\ {-3x^2 + 1 \over x(x^2 + 1)} & = {1 \over x} + {(-4)x + (0) \over x^2 + 1} \\ & = {1 \over x} + {-4x \over x^2 + 1} \\ & = {1 \over x} - {4x \over x^2 + 1} \\ \\ {4x^5 + 2x^4 + 3x^3 - x^2 -x + 1 \over x(x^2 + 1)} & = 4x^2 + 2x - 1 + {1 \over x} - {4x \over x^2 + 1} \end{align}

\begin{align} x(2x - 5) & = 2x^2 - 5x \end{align} $$ \require{enclose} \begin{array}{rll} 3 \phantom{00000000000}\\ 2x^2 - 5x \enclose{longdiv}{ 6x^2 - 21x + 25 \phantom{0}}\kern-.2ex \\ -\underline{( 6x^2 - 15x ){\phantom{0000.}}} \\ -6x + 25\phantom{0} \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ {6x^2 - 21x + 25 \over x(2x - 5)} & = 3 + {-6x + 25 \over x(2x - 5)} \\ \\ {-6x + 25 \over x(2x - 5)} & = {A \over x} + {B \over 2x - 5} \\ & = {A(2x - 5) \over x(2x - 5)} + {B(x) \over x(2x - 5)} \\ & = {A(2x - 5) + B(x) \over x(2x - 5)} \\ \\ \text{Comparing } & \text{the numerator,} \\ -6x + 25 & = A(2x - 5) + B(x) \\ \\ \text{Let } & x = 2.5, \\ -6(2.5) + 25 & = A[2(2.5) - 5] + B(2.5) \\ 10 & = A(0) + 2.5B \\ 10 & = 2.5B \\ {10 \over 2.5} & = B \\ 4 & = B \\ \\ \text{Let } & x = 0, \\ -6(0) + 25 & = A[2(0) - 5] + (4)(0) \\ 25 & = A(-5) + 0 \\ 25 & = -5A \\ {25 \over -5} & = A \\ -5 & = A \\ \\ {-6x + 25 \over x(2x - 5)} & = {-5 \over x} + {4 \over 2x - 5} \\ \\ \therefore {6x^2 - 21x + 25 \over x(2x - 5)} & = 3 + {-5 \over x} + {4 \over 2x - 5} \\ \\ A = 3, B & = -5, C = 4 \end{align}

$$ \require{enclose} \begin{array}{rll} x + 3 \phantom{00000000000}\\ 4x^2 + 0x - 1 \enclose{longdiv}{ 4x^3 + 12x^2 - x - 4 \phantom{0}}\kern-.2ex \\ -\underline{( 4x^3 + 0x^2 - x ){\phantom{0000.}}} \\ 12x^2 + 0x - 4\phantom{0} \\ -\underline{( 12x^2 + 0x - 3 ){\phantom{}}} \\ -1\phantom{0} \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ {4x^3 + 12x^2 - x - 4 \over 4x^2 - 1} & = x + 3 + {3x - 1 \over 4x^2 - 1} \\ & = x + 3 + {- 1 \over (2x - 1)(2x + 1)} \\ \\ {- 1 \over (2x - 1)(2x + 1)} & = {A \over 2x - 1} + {B \over 2x + 1} \\ & = {A(2x + 1) \over (2x - 1)(2x + 1)} + {B(2x - 1) \over (2x - 1)(2x + 1)} \\ & = {A(2x + 1) + B(2x - 1) \over (2x - 1)(2x + 1)} \\ \\ \\ \text{Comparing } & \text{the numerator,} \\ - 1 & = A(2x + 1) + B(2x - 1) \\ \\ \text{Let } & x = -0.5, \\ - 1 & = A[2(-0.5) + 1] + B[2(-0.5) - 1] \\ -1 & = A(0) + B(-2) \\ -1 & = -2B \\ {-1 \over -2} & = B \\ {1 \over 2} & = B \\ \\ \text{Let } & x = 0.5, \\ -1 & = A[2(0.5) + 1] + {1 \over 2}[2(0.5) - 1] \\ -1 & = A(2) + {1 \over 2}(0) \\ -1 & = 2A \\ {-1 \over 2} & = A \\ -{1 \over 2} & = A \\ \\ {- 1 \over (2x - 1)(2x + 1)} & = {-{1 \over 2} \over 2x - 1} + {{1 \over 2} \over 2x + 1} \\ \\ \therefore {4x^3 + 12x^2 - x - 4 \over 4x^2 - 1} & = x + 3 + {-{1 \over 2} \over 2x - 1} + {{1 \over 2} \over 2x + 1} \\ \\ A = 1, B = 3, C & = -{1 \over 2}, D = {1 \over 2} \end{align}

(i)

\begin{align} \text{Height} & = {\text{Volume} \over \text{Cross-sectional area}} \\ \\ & = {3x^2 + 8x + 1 \over x^2 + 2x + 1} \\ \\ & = {3x^2 + 8x + 1 \over (x + 1)^2} \end{align}

(ii)

$$ \require{enclose} \begin{array}{rll} 3 \phantom{000000000}\\ x^2 + 2x + 1 \enclose{longdiv}{ 3x^2 + 8x + 1 \phantom{0}}\kern-.2ex \\ -\underline{( 3x^2 + 6x + 3 ){\phantom{.}}} \\ 2x - 2\phantom{0} \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ {3x^2 + 8x + 1 \over (x + 1)^2} & = 3 + {2x - 2 \over (x + 1)^2} \\ \\ {2x - 2 \over (x + 1)^2} & = {A \over x + 1} + {B \over (x + 1)^2} \\ & = {A(x + 1) \over (x + 1)^2} + {B \over (x + 1)^2} \\ & = {A(x + 1) + B \over (x + 1)^2} \\ \\ \text{Comparing } & \text{the numerator,} \\ 2x - 2 & = A(x + 1) + B \\ \\ \text{Let } & x = -1, \\ 2(-1) - 2 & = A(-1 + 1) + B \\ -4 & = A(0) + B \\ -4 & = B \\ \\ \text{Let } & x = 0, \\ 2(0) - 2 & = A(0 + 1) + (-4) \\ -2 & = A(1) - 4 \\ -2 + 4 & = A \\ 2 & = A \\ \\ {2x - 2 \over (x + 1)^2} & = {2 \over x + 1} + {-4 \over (x + 1)^2} \\ & = {2 \over x + 1} - {4 \over (x + 1)^2} \\ \\ \therefore {3x^2 + 8x + 1 \over (x + 1)^2} & = 3 + {2 \over x + 1} - {4 \over (x + 1)^2} \end{align}

(i)

\begin{align} \text{Let } f(x) & = x^3 - 3x^2 + 4 \\ \\ f(-1) & = (-1)^3 - 3(-1)^2 + 4 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } x + 1 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 - 4x + 4 \phantom{000000}\\ x + 1 \enclose{longdiv}{ x^3 - 3x^2 + 0x + 4 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 + x^2 ){\phantom{00000000.}}} \\ -4x^2 + 0x + 4\phantom{0} \\ -\underline{( -4x^2 - 4x ){\phantom{000.}}} \\ 4x + 4 \phantom{0} \\ -\underline{( 4x + 4 ){\phantom{.}}} \\ 0\phantom{0} \\ \end{array} $$ \begin{align} \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder,} \\ f(x) & = (x + 1)(x^2 - 4x + 4) \\ & = (x + 1)[x^2 - 2(x)(2) + 2^2] \\ & = (x + 1)(x - 2)^2 \end{align}

(ii)

$$ \require{enclose} \begin{array}{rll} x^2 + 3x + 9 \phantom{000000000000000000}\\ x^3 - 3x^2 + 0x + 4 \enclose{longdiv}{ x^5 + 0x^4 + 0x^3 - 36x^2 + 53x + 18 \phantom{0}}\kern-.2ex \\ -\underline{( x^5 - 3x^4 + 0x^3 + 4x^2 ){\phantom{0000000000.}}} \\ 3x^4 + 0x^3 - 40x^2 + 53x + 18 \phantom{0.} \\ -\underline{( 3x^4 - 9x^3 + 0x^2 + 12x ){\phantom{00000.}}} \\ 9x^3 - 40x^2 + 41x + 18 \phantom{0} \\ -\underline{( 9x^3 - 27x^2 + 0x + 36 ){\phantom{.}}} \\ -13x^2 + 41x - 18 \phantom{0} \\ \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ {x^5 - 36x^2 + 53x + 18 \over x^3 - 3x^2 + 4} & = x^2 + 3x + 9 + {-13x^2 + 41x -18 \over x^3 - 3x^2 + 4} \\ \\ {-13x^2 + 41x -18 \over x^3 - 3x^2 + 4} & = {-13x^2 + 41x -18 \over (x + 1)(x - 2)^2} \\ & = {A \over x + 1} + {B \over x - 2} + {C \over (x - 2)^2} \\ & = {A(x - 2)^2 \over (x + 1)(x - 2)^2} + {B(x + 1)(x - 2) \over (x + 1)(x - 2)^2} + {C(x + 1) \over (x + 1)(x - 2)^2} \\ & = {A(x - 2)^2 + B(x + 1)(x - 2) + C(x + 1) \over (x + 1)(x - 2)^2} \\ \\ \text{Comparing } & \text{the numerator,} \\ -13x^2 + 41x -18 & = A(x - 2)^2 + B(x + 1)(x - 2) + C(x + 1) \\ \\ \text{Let } & x = 2, \\ -13(2)^2 + 41(2) - 18 & = A(2 - 2)^2 + B(2 + 1)(2 - 2) + C(2 + 1) \\ 12 & = A(0)^2 + B(3)(0) + C(3) \\ 12 & = 3C \\ {12 \over 3} & = C \\ 4 & = C \\ \\ \text{Let } & x = -1, \\ -13(-1)^2 + 41(-1) - 18 & = A(-1 - 2)^2 + B(-1 + 1)(-1 - 2) + (4)(-1 + 1) \\ -72 & = A(-3)^2 + B(0)(-3) + (4)(0) \\ -72 & = A(9) \\ -72 & = 9A \\ {-72 \over 9} & = A \\ -8 & = A \\ \\ \text{Let } & x = 0, \\ -13(0)^2 + 41(0) - 18 & = (-8)(0 - 2)^2 + B(0 + 1)(0 - 2) + (4)(0 + 1) \\ -18 & = (-8)(-2)^2 + B(1)(-2) + (4)(1) \\ -18 & = (-8)(4) - 2B + 4 \\ -18 & = -32 - 2B + 4 \\ 2B & = -32 + 4 + 18 \\ & = -10 \\ B & = {-10 \over 2} \\ & = -5 \\ \\ {-13x^2 + 41x -18 \over x^3 - 3x^2 + 4} & = {-8 \over x + 1} + {-5 \over x - 2} + {4 \over (x - 2)^2} \\ \\ \therefore {x^5 - 36x^2 + 53x + 18 \over x^3 - 3x^2 + 4} & = x^2 + 3x + 9 + {-8 \over x + 1} + {-5 \over x - 2} + {4 \over (x - 2)^2} \\ & = x^2 + 3x + 9 - {8 \over x + 1} - {5 \over x - 2} + {4 \over (x - 2)^2} \end{align}

\begin{align} {x + 8 \over (x - 4)(x + 2)} & = {A \over x - 4} + {B \over x + 2} \\ & = {A(x + 2) \over (x - 4)(x + 2)} + {B(x - 4) \over (x - 4)(x + 2)} \\ & = {A(x + 2) + B(x - 4) \over (x - 4)(x + 2)} \\ \\ \text{Comparing } & \text{the numerator,} \\ x + 8 & = A(x + 2) + B(x - 4) \\ \\ \text{Let } & x = -2, \\ (-2) + 8 & = A(-2 + 2) + B(-2 - 4) \\ 6 & = A(0) + B(-6) \\ 6 & = -6B \\ {6 \over -6} & = B \\ -1 & = B \\ \\ \text{Let } & x = 4, \\ (4) + 8 & = A(4 + 2) + (-1)(4 - 4) \\ 12 & = A(6) + (-1)(0) \\ 12 & = 6A \\ {12 \over 6} & = A \\ 2 & = A \\ \\ \therefore {x + 8 \over (x - 4)(x + 2)} & = {2 \over x - 4} + {-1 \over x + 2} \\ & = {2 \over x - 4} - {1 \over x + 2} \end{align}

(i)

\begin{align} {2x + 16 \over (x - 4)(x + 2)} + {2 \over x + 2} & = 1 \\ {2(x + 8) \over (x - 4)(x + 2)} + {2 \over x + 2} & = 1 \\ 2 \left[{x + 8 \over (x - 4)(x + 2)}\right] + {2 \over x + 2} & = 1 \\ \\ [\text{Apply result form first part}] \phantom{00000000} 2 \left( {2 \over x - 4} - {1 \over x + 2} \right) + {2 \over x + 2} & = 1 \\ {4 \over x - 4} - {2 \over x + 2} + {2 \over x + 2} & = 1 \\ {4 \over x - 4} & = 1 \\ 4 & = x - 4 \\ 4 + 4 & = x \\ 8 & = x \end{align}

(ii)

\begin{align} {x + 8 \over (x - 4)(x + 2)} & = {2 \over x - 4} - {1 \over x + 2} \\ \\ \text{Let } & x = x^2, \\ {x^2 + 8 \over (x^2 - 4)(x^2 + 2)} & = {2 \over x^2 - 4} - {1 \over x^2 + 2} \\ & = {2 \over (x + 2)(x - 2)} - {1 \over x^2 + 2} \\ \\ {2 \over (x + 2)(x - 2)} & = {A \over x + 2} + {B \over x - 2} \\ & = {A(x - 2) \over (x + 2)(x - 2)} + {B(x + 2) \over (x + 2)(x - 2)} \\ & = {A(x - 2) + B(x + 2) \over (x + 2)(x - 2)} \\ \\ \text{Comparing } & \text{the numerator,} \\ 2 & = A(x - 2) + B(x + 2) \\ \\ \text{Let } & x = 2, \\ 2 & = A(2 - 2) + B(2 + 2) \\ 2 & = A(0) + B(4) \\ 2 & = 4B \\ {2 \over 4} & = B \\ {1 \over 2} & = B \\ \\ \text{Let } & x = -2, \\ 2 & = A(-2 -2) + {1 \over 2}(-2 + 2) \\ 2 & = A(-4) + {1 \over 2}(0) \\ 2 & = -4A \\ {2 \over -4} & = A \\ -{1 \over 2} & = A \\ \\ {2 \over (x + 2)(x - 2)} & = {-{1 \over 2} \over x + 2} + {{1 \over 2} \over x - 2} \\ & = -{1 \over 2(x + 2)} + {1 \over 2(x - 2)} \\ \\ \therefore {x^2 + 8 \over (x^2 - 4)(x^2 + 2)} & = {2 \over (x + 2)(x - 2)} - {1 \over x^2 + 2} \\ & = -{1 \over 2(x + 2)} + {1 \over 2(x - 2)} - {1 \over x^2 + 2} \end{align}

(i)

\begin{align} {100x + 250 \over x^2 + 5x} & = {100x + 250 \over x(x + 5)} \\ & = {A \over x} + {B \over x + 5} \\ & = {A(x + 5) \over x(x + 5)} + {B(x) \over x(x + 5)} \\ & = {A(x + 5) + B(x) \over x(x + 5)} \\ \\ \text{Comparing } & \text{the numerator,} \\ 100x + 250 & = A(x + 5) + B(x) \\ \\ \text{Let } & x = - 5, \\ 100(-5) + 250 & = A(-5 + 5) + B(-5) \\ -250 & = A(0) - 5B \\ -250 & = -5B \\ {-250 \over -5} & = B \\ 50 & = B \\ \\ \text{Let } & x = 0, \\ 100(0) + 250 & = A(0 + 5) + (50)(0) \\ 250 & = A(5) \\ 250 & = 5A \\ {250 \over 5} & = A \\ 50 & = A \\ \\ \therefore {100x + 250 \over x^2 + 5x} & = {50 \over x} + {50 \over x + 5} \end{align}

(ii)

\begin{align} \text{Distance} & = \text{Speed} \times \text{Time} \\ \\ \text{For } & \text{initial journey,} \\ \text{Distance} & = x \times {50 \over x} \\ & = 50 \text{ km} \end{align}

(iii)

${50 \over x}$ is the time taken for the initial journey while ${50 \over x + 5}$ is the time taken for the return journey.

(i)

\begin{align} \text{Since } & a^3 - b^3 = (a - b)(a^2 + ab + b^2), \\ x^3 - 8 & = (x)^3 - (2)^3 \\ & = (x - 2)[x^2 + (x)(2) + 2^2] \\ & = (x - 2)(x^2 + 2x + 4) \end{align}

(ii)

\begin{align} {x^2 + 4 \over x^3 - 8} & = {x^2 + 4 \over (x - 2)(x^2 + 2x + 4)} \\ & = {A \over x - 2} + {Bx + C \over x^2 + 2x + 4} \\ & = {A(x^2 + 2x + 4) + (Bx + C)(x - 2) \over (x - 2)(x^2 + 2x + 4)} \\ \\ \text{Comparing } & \text{the numerator,} \\ x^2 + 4 & = A(x^2 + 2x + 4) + (Bx + C)(x - 2) \\ \\ \text{Let } & x = 2, \\ (2)^2 + 4 & = A[(2)^2 + 2(2) + 4] + [B(2) + C](2 - 2) \\ 8 & = A(12) + (2B + C)(0) \\ 8 & = 12A \\ {8 \over 12} & = A \\ {2 \over 3} & = A \\ \\ \text{Let } & x = 0, \\ (0)^2 + 4 & = {2 \over 3}[(0)^2 + 2(0) + 4] + [B(0) + C](0 - 2) \\ 4 & = {2 \over 3}(4) + (C)(-2) \\ 4 & = {8 \over 3} - 2C \\ 2C & = {8 \over 3} - 4 \\ & = -{4 \over 3} \\ C & = -{4 \over 3} \div 2 \\ & = -{2 \over 3} \\ \\ \text{Let } & x = 1, \\ (1)^2 + 4 & = {2 \over 3}[(1)^2 + 2(1) + 4] + \left[B(1) + \left(-{2 \over 3}\right)\right])(1 - 2) \\ 5 & = {2 \over 3}(7) + \left(B - {2 \over 3}\right)(-1) \\ 5 & = {14 \over 3} - B + {2 \over 3} \\ B & = {14 \over 3} + {2 \over 3} - 5 \\ & = {1 \over 3} \\ \\ \therefore {x^2 + 4 \over x^3 - 8} & = {A \over x - 2} + {Bx + C \over x^2 + 2x + 4} \\ & = {{2 \over 3} \over x - 2} + {({1 \over 3})x + (-{2 \over 3}) \over x^2 + 2x + 4} \\ & = {2 \over 3(x - 2)} + {{1 \over 3}x - {2 \over 3} \over x^2 + 2x + 4} \\ & = {2 \over 3(x - 2)} + {{1 \over 3}(x - 2) \over x^2 + 2x + 4} \\ & = {2 \over 3(x - 2)} + {x - 2 \over 3(x^2 + 2x + 4)} \end{align}

(i)

\begin{align} {1 \over (x + 1)(x + 2)} & = {A \over x + 1} + {B \over x + 2} \\ & = {A(x + 2) \over (x + 1)(x + 2)} + {B(x + 1) \over (x + 1)(x + 2)} \\ & = {A(x + 2) + B(x + 1) \over (x + 1)(x + 2)} \\ \\ 1 & = A(x + 2) + B(x + 1) \\ \\ \text{Let } & x = -2, \\ 1 & = A(0) + B(-2 + 1) \\ 1 & = 0 + B(-1) \\ 1 & = -B \\ -1 & = B \\ \\ \text{Let } & x = -1, \\ 1 & = A(-1 + 2) + B(0) \\ 1 & = A(1) + 0 \\ 1 & = A \\ \\ {1 \over (x + 1)(x + 2)} & = {1 \over x + 1} + {-1 \over x + 2} \\ & = {1 \over x + 1} - {1 \over x + 2} \end{align}

(ii)

\begin{align} \text{From (i), } {1 \over (x + 1)(x + 2)} & = {1 \over x + 1} - {1 \over x + 2} \\ \\ {1 \over 2(3)} = {1 \over (1 + 1)(1 + 2)} & = {1 \over 1 + 1} - {1 \over 1 + 2} \\ & = {1 \over 2} - {1 \over 3} \end{align}

(iii)

\begin{align} \require{cancel} {1 \over 3(4)} = {1 \over (2 + 1)(2 + 2)} & = {1 \over 2 + 1} - {1 \over 2 + 2} \\ & = {1 \over 3} - {1 \over 4} \\ \\ \implies {1 \over 4(5)} & = {1 \over 4} - {1 \over 5} \\ \\ {1 \over 101(102)} & = {1 \over 101} - {1 \over 102} \\ \\ \\ \therefore \text{Sum of fractions} & = {1 \over 2} - \cancel{{1 \over 3}} \\ & + \cancel{{1 \over 3}} - \cancel{{1 \over 4}} \\ & + \cancel{{1 \over 4}} - \cancel{{1 \over 5}} \\ & + \phantom{0} ... \\ & + \phantom{0} ... \\ & + \phantom{0} ... \\ & + \cancel{{1 \over 101}} - {1 \over 102} \\ & = {1 \over 2} - {1 \over 102} \\ & = {25 \over 51} \end{align}

(i)

\begin{align} {2 \over x - 1} + {3x + 1 \over x^2 + 4} & = {2(x^2 + 4) \over (x - 1)(x^2 + 4)} + {(x - 1)(3x + 1) \over (x - 1)(x^2 + 4)} \\ & = {2(x^2 + 4) + (x - 1)(3x + 1) \over (x - 1)(x^2 + 4)} \\ & = {2x^2 + 8 + 3x^2 + x - 3x - 1 \over (x - 1)(x^2 + 4)} \\ & = {5x^2 - 2x + 7 \over (x - 1)(x^2 + 4)} \end{align}

(ii)

\begin{align} {5x^2 - 3x + 8 \over (x - 1)(x^2 + 4)} & = {A \over x - 1} + {Bx + C \over x^2 + 4} \\ & = {A(x^2 + 4) \over (x - 1)(x^2 + 4)} + {(x - 1)(Bx + C) \over (x - 1)(x^2 + 4)} \\ & = {A(x^2 + 4) + (x - 1)(Bx + C) \over (x - 1)(x^2 + 4)} \\ \\ 5x^2 - 3x + 8 & = A(x^2 + 4) + (x - 1)(Bx + C) \\ \\ \text{Let } & x = 1, \\ 5(1)^2 - 3(1) + 8 & = A(1^2 + 4) + 0 \\ 10 & = A(5) \\ 10 & = 5A \\ {10 \over 5} & = A \\ 2 & = A \\ \\ 5x^2 - 3x + 8 & = 2(x^2 + 4) + (x - 1)(Bx + C) \\ \\ \text{Let } & x = 0, \\ 0 - 0 + 8 & = 2(0^2 + 4) + (0 - 1)(0 + C) \\ 8 & = 8 + (-1)(C) \\ 8 & = 8 - C \\ 0 & = -C \\ 0 & = C \\ \\ 5x^2 - 3x + 8 & = 2(x^2 + 4) + (x - 1)(Bx) \\ \\ \text{Let } & x = 2, \\ 5(2)^2 - 3(2) + 8 & = 2(2^2 + 4) + (2 - 1)(2B) \\ 22 & = 16 + (1)(2B) \\ 22 & = 16 + 2B \\ 6 & = 2B \\ {6 \over 2} & = B \\ 3 & = B \\ \\ \therefore {5x^2 - 3x + 8 \over (x - 1)(x^2 + 4)} & = {2 \over x - 1} + {3x \over x^2 + 4} \end{align}