(a) (i) Multiply out and simplify $(x + 5)^2$ - Junior Cycle Mathematics - Question 11 - 2016

Now, we compute $(x + 5)^2 - (x - 5)^2$. Using the difference of squares formula, we have:

$$(a^2 - b^2 = (a - b)(a + b)) ext{ where } a = (x + 5) ext{ and } b = (x - 5).$$

So,

$$(x + 5)^2 - (x - 5)^2 = [(x + 5) - (x - 5)][(x + 5) + (x - 5)]$$

This simplifies to:

$$(10)(2x) = 20x.$$

Since $20x$ can be factored into $4(5x)$, we conclude that:

$$20x ext{ is divisible by } 4.$$

For part (b):

(b) (i) Factorising $25x^2 - 49n^2$, we recognize it as a difference of squares, which factors as:

$$(a^2 - b^2 = (a - b)(a + b)) ext{ where } a = 5x ext{ and } b = 7n.$$

Hence,

$$25x^2 - 49n^2 = (5x - 7n)(5x + 7n).$$

(b) (ii) For $2x^2 - 9x - 18$, we first find two numbers that multiply to $(2 * -18 = -36)$ and add to $-9$. The numbers are $-12$ and $3$.

This allows us to factor by grouping:

$$2x^2 - 12x + 3x - 18 = 2x(x - 6) + 3(x - 6) = (2x + 3)(x - 6).$$