a) (i) Factorise $n^2 - 1$ - Junior Cycle Mathematics - Question 12 - 2015

Let the two consecutive odd numbers be represented as:

$$(n - 1) ext{ and } (n + 1)$$

The product is given by:

$$(n - 1)(n + 1) = 399$$

Expanding this gives:

$$n^2 - 1 = 399$$

Thus,

$$n^2 = 400$$

Taking the square root of both sides, we get:

$$n = 20$$

Substituting back:

The two numbers are:

$$19 ext{ and } 21$$

Using polynomial long division, we start by dividing the leading term of the dividend by the leading term of the divisor:

1. Divide:

   $$$$

rac{x^3}{x} = x^2

$$Multiply:$$

(x + 3)(x^2) = x^3 + 3x^2

$$Subtract:$$

(x^3 + 5x^2 - 29x - 105) - (x^3 + 3x^2) = 2x^2 - 29x - 105

2. Next, repeat with $2x^2$: $$ rac{2x^2}{x} = 2x

Multiply:

$$(x + 3)(2x) = 2x^2 + 6x$$

Subtract:

$$(2x^2 - 29x - 105) - (2x^2 + 6x) = -35x - 105$$

3. Repeat with $-35x$:

   $$$$

rac{-35x}{x} = -35

$$Multiply:$$

(x + 3)(-35) = -35x - 105

$$Subtract:$$

(-35x - 105) - (-35x - 105) = 0

$$Thus, the division results in:$$

x^2 + 2x - 35

$$$$