12 (a) Factorise $n^2 - 11n + 18$ - Junior Cycle Mathematics - Question 12 - 2017

First, we can rearrange the expression:

$wy - y + w - 1 = y(w - 1) + 1(w - 1)$.

Now we can factor by grouping:

$(w - 1)(y + 1)$.

Substituting $x = 4$:

First compute $3(4) - 2 = 12 - 2 = 10$ and $6(4) - 12 = 24 - 12 = 12$.

So the expression becomes:

$\frac{5}{10} - \frac{7}{12} = \frac{1}{2} - \frac{7}{12}$.

Finding a common denominator (which is $12$):

$\frac{6}{12} - \frac{7}{12} = -\frac{1}{12}$.

Start by factorising the numerator and the denominator separately. The numerator can be expressed as a difference of squares:

$4e^2 - 9 = (2e - 3)(2e + 3).$

Next, finding factors of $2e^2 + 3e - 9$ can lead us to expressions such as $(2e - 3)(e + 3)$. Thus, the full simplification is:

$\frac{(2e - 3)(2e + 3)}{(2e - 3)(e + 3)} = \frac{2e + 3}{e + 3}$, assuming $2e - 3 \neq 0$.

To find the coefficients $a$, $b$, and $c$ in the expression $ax^2 + bx + c$, first, set the area equal to:

$2x^3 - 13x^2 + 25x - 12$.

Matching coefficients gives:

$a = 2$,

$b = -13$,

$c = -12$.