Given P(x) = bx3 − x2 + cx − b, if (x+1) and (x-2) are factors of P(x), find the values of b and c. Show that (x-4) is not a factor and find the remaining factor.

1) a)
Let the given expression be $P(x) = bx^3 - x^2 + cx - b$.
Since $(x+1)$ is a factor, $P(-1) = 0$.

$$b(-1)^3 - (-1)^2 + c(-1) - b = 0$$ $$-b - 1 - c - b = 0$$ $$-2b - c - 1 = 0$$ $$2b + c = -1 \quad (1)$$

Since $(x-2)$ is a factor, $P(2) = 0$.

$$b(2)^3 - (2)^2 + c(2) - b = 0$$ $$8b - 4 + 2c - b = 0$$ $$7b + 2c - 4 = 0$$ $$7b + 2c = 4 \quad (2)$$

From equation (1), $c = -1 - 2b$. Substitute this into equation (2):

$$7b + 2(-1 - 2b) = 4$$ $$7b - 2 - 4b = 4$$ $$3b - 2 = 4$$ $$3b = 6$$ $$b = 2$$

Now substitute $b=2$ back into $c = -1 - 2b$:

$$c = -1 - 2(2)$$ $$c = -1 - 4$$ $$c = -5$$

The values are $\boxed{b=2, c=-5}$.

b)
Substitute the values of $b$ and $c$ into the expression:

$$P(x) = 2x^3 - x^2 - 5x - 2$$

The expression is $\boxed{2x^3 - x^2 - 5x - 2}$.

c)
To show that $(x-4)$ is not a factor, we evaluate $P(4)$. If $P(4) \neq 0$, then $(x-4)$ is not a factor.

$$P(4) = 2(4)^3 - (4)^2 - 5(4) - 2$$ $$P(4) = 2(64) - 16 - 20 - 2$$ $$P(4) = 128 - 16 - 20 - 2$$ $$P(4) = 112 - 20 - 2$$ $$P(4) = 92 - 2$$ $$P(4) = 90$$

Since $P(4) = 90 \neq 0$, $(x-4)$ is not a factor of the expression.

d)
We know that $(x+1)$ and $(x-2)$ are factors. Their product is:

$$(x+1)(x-2) = x^2 - 2x + x - 2 = x^2 - x - 2$$

Now, we divide $P(x) = 2x^3 - x^2 - 5x - 2$ by $(x^2 - x - 2)$ to find the remaining factor.
Using polynomial long division:

$$\begin{array}{c|cc cc} \multicolumn{2}{r}{2x} & +1 \\ \cline{2-5} x^2-x-2 & 2x^3 & -x^2 & -5x & -2 \\ \multicolumn{2}{r}{-(2x^3} & -2x^2 & -4x) \\ \cline{2-4} \multicolumn{2}{r}{0} & x^2 & -x & -2 \\ \multicolumn{2}{r}{} & -(x^2 & -x & -2) \\ \cline{3-5} \multicolumn{2}{r}{} & 0 & 0 & 0 \\ \end{array}$$

The remaining factor is $(2x+1)$.
To solve for $x$ when the expression is equated to zero:

$$(x+1)(x-2)(2x+1) = 0$$

Setting each factor to zero:

$$x+1 = 0 \implies x = -1$$ $$x-2 = 0 \implies x = 2$$ $$2x+1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}$$

The remaining factor is $\boxed{2x+1}$ and the values of $x$ are $\boxed{-1, 2, -\frac{1}{2}}$.

2)
Express $f(x) = \frac{x}{x^2 - 4}$ in partial fractions.
First, factor the denominator: $x^2 - 4 = (x-2)(x+2)$.
So, $f(x) = \frac{x}{(x-2)(x+2)}$.
We set up the partial fraction decomposition:

$$\frac{x}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}$$

Multiply both sides by $(x-2)(x+2)$:

$$x = A(x+2) + B(x-2)$$

To find $A$, let $x=2$:

$$2 = A(2+2) + B(2-2)$$ $$2 = 4A + 0$$ $$4A = 2$$ $$A = \frac{2}{4} = \frac{1}{2}$$

To find $B$, let $x=-2$:

$$-2 = A(-2+2) + B(-2-2)$$ $$-2 = 0 + B(-4)$$ $$-2 = -4B$$ $$B = \frac{-2}{-4} = \frac{1}{2}$$

Substitute the values of $A$ and $B$ back into the decomposition:

$$f(x) = \frac{\frac{1}{2}}{x-2} + \frac{\frac{1}{2}}{x+2}$$ $$f(x) = \frac{1}{2(x-2)} + \frac{1}{2(x+2)}$$

The partial fraction decomposition is $\boxed{\frac{1}{2(x-2)} + \frac{1}{2(x+2)}}$.