Factorise fully each of the following expressions - Junior Cycle Mathematics - Question 10 - 2014

For the expression $rc - sc + 2rd - 2sd$, we can group the terms as follows:

$$(rc - sc) + (2rd - 2sd)$$

Next, we factor out common terms from each group:

$$= c(r - s) + 2d(r - s)$$

Now, we see that $(r - s)$ is a common factor:

$$= (c + 2d)(r - s)$$

The expression $x² - 16$ is a difference of squares. We can express it as:

$$(x + 4)(x - 4)$$

To factorise the quadratic $x² - 5x + 6$, we look for two numbers that multiply to +6 and add to -5. These numbers are -2 and -3. Thus, we factor the expression as:

$$(x - 3)(x - 2)$$

Using the factors from (b)(i), we set:

$$(x - 3)(x - 2) = 0$$

This gives two equations:

$$x - 3 = 0 \\ ext{or} \\ x - 2 = 0$$

Thus, the solutions are:

$$x = 3 \\ ext{and} \\ x = 2$$

Let's verify $x = 3$. We substitute into the original equation:

$$3^2 - 5(3) + 6 = 0$$

Calculating:

$$9 - 15 + 6 = 0$$

This simplifies to:

$$0 = 0,$$

which confirms that $x = 3$ is indeed a solution.