Express $\frac{8x^3 + 3x + 1}{x(2x + 1)^2}$ as a sum of a polynomial and a proper fraction. Hence, express $\frac{8x^3 + 3x + 1}{x(2x + 1)^2}$ as a sum of a polynomial and a partial fraction. Space for Q8(b)

1. Expand the denominator to prepare for long division.

$x(2x + 1)^2 = x(4x^2 + 4x + 1) = 4x^3 + 4x^2 + x$

2. Perform polynomial long division to find the quotient and remainder.

$\frac{8x^3 + 3x + 1}{4x^3 + 4x^2 + x}$

Quotient $= 2$

Remainder $= (8x^3 + 3x + 1) - 2(4x^3 + 4x^2 + x)$
$= 8x^3 + 3x + 1 - 8x^3 - 8x^2 - 2x$
$= -8x^2 + x + 1$

3. Express as a sum of a polynomial and a proper fraction.

$\frac{8x^3 + 3x + 1}{x(2x + 1)^2} = 2 + \frac{-8x^2 + x + 1}{x(2x + 1)^2}$

1. Perform long division to express the given expression as a polynomial and a proper fraction (from part a).

Denominator: $x(2x + 1)^2 = x(4x^2 + 4x + 1) = 4x^3 + 4x^2 + x$

By long division:
$8x^3 + 3x + 1 = 2(4x^3 + 4x^2 + x) - 8x^2 + x + 1$

$\frac{8x^3 + 3x + 1}{x(2x + 1)^2} = 2 + \frac{-8x^2 + x + 1}{x(2x + 1)^2}$

2. Set up the partial fraction decomposition for the proper fraction.

Let $\frac{-8x^2 + x + 1}{x(2x + 1)^2} = \frac{A}{x} + \frac{B}{2x + 1} + \frac{C}{(2x + 1)^2}$

$-8x^2 + x + 1 = A(2x + 1)^2 + Bx(2x + 1) + Cx$

3. Solve for the constants A, B, and C.

Substitute $x = 0$:
$1 = A(1)^2 \implies A = 1$

Substitute $x = -\frac{1}{2}$:
$-8(-\frac{1}{2})^2 + (-\frac{1}{2}) + 1 = C(-\frac{1}{2})$
$-2 - \frac{1}{2} + 1 = -\frac{1}{2}C -\frac{3}{2} = -\frac{1}{2}C \implies C = 3$

Compare coefficients of $x^2$:
$-8 = 4A + 2B -8 = 4(1) + 2B 2B = -12 \implies B = -6$

4. Write the final expression.

$\frac{8x^3 + 3x + 1}{x(2x + 1)^2} = 2 + \frac{1}{x} - \frac{6}{2x + 1} + \frac{3}{(2x + 1)^2}$