Given f(x) = 3 − ax + bx2 − x3, find the values of a and b if f(x) divided by (x+2) gives a remainder of 45 and (3-x) is a factor of f(x). Also decompose (5x)/((x+3)(x-2)) into partial fractions.

Given $f(x) = 3 - ax + bx^2 - x^3$.
When $f(x)$ is divided by $(x+2)$, the remainder is 45. This means $f(-2) = 45$.

$$f(-2) = 3 - a(-2) + b(-2)^2 - (-2)^3 = 45$$ $$3 + 2a + 4b - (-8) = 45$$ $$3 + 2a + 4b + 8 = 45$$ $$11 + 2a + 4b = 45$$ $$2a + 4b = 34$$

Divide by 2:

$$a + 2b = 17 \quad \text{(Equation 1)}$$

Given that $(3-x)$ is a factor of $f(x)$. This means $f(3) = 0$.

$$f(3) = 3 - a(3) + b(3)^2 - (3)^3 = 0$$ $$3 - 3a + 9b - 27 = 0$$ $$-24 - 3a + 9b = 0$$ $$-3a + 9b = 24$$

Divide by 3:

$$-a + 3b = 8 \quad \text{(Equation 2)}$$

Add Equation 1 and Equation 2:

$$(a + 2b) + (-a + 3b) = 17 + 8$$ $$5b = 25$$ $$b = 5$$

Substitute $b=5$ into Equation 1:

$$a + 2(5) = 17$$ $$a + 10 = 17$$ $$a = 7$$

The values of the real constants are $\boxed{a=7, b=5}$.

1. (ii)

$$\frac{5x}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$$

Multiply both sides by $(x+3)(x-2)$:

$$5x = A(x-2) + B(x+3)$$

$$5(-3) = A(-3-2) + B(-3+3)$$ $$-15 = A(-5) + B(0)$$ $$-15 = -5A$$ $$A = 3$$

$$5(2) = A(2-2) + B(2+3)$$ $$10 = A(0) + B(5)$$ $$10 = 5B$$ $$B = 2$$