Perhaps you're unsure of how to do the ' hence' for ii) because you did i) this way instead of long division :

Let f(x) = x³ + x² + 3x + 3

When divided by (x + 1), remainder = f(-1)
= (-1)³ + (-1)² + 3(-1) + 3
= -1 + 1 - 3 + 3
= 0


For ii) this is how you continue :

Since f(-1) = 0, by the factor theorem, (x + 1) is a factor of f(x)

What you want to do now is to factor (x + 1) out.

Notice that we can do so via the following :

x³ + x² + 3x + 3

= x²(x + 1) + 3(x + 1)

= (x + 1)(x² + 3)


Now you're ready to express (5x² + x + 4) / (x³ + x² + 3x + 3) in partial fractions.


(5x² + x + 4) / (x³ + x² + 3x + 3)

= (5x² + x + 4) / [(x + 1)(x² + 3)]


Apply your usual partial fractions method accordingly.

Here's a manual way to do the partial fractions :


(5x² + x + 4) / [(x + 1)(x² + 3)]

= (x² + 3 + 4x² + x + 1) / [(x + 1)(x² + 3)]

= (x² + 3 + x² + 3 + 3x² + x - 2) / [(x + 1)(x² + 3)]

= (2(x² + 3) + (3x - 2)(x + 1)) / [(x + 1)(x² + 3)]


= 2 / (x + 1) + (3x - 2)/(x² + 3)