Algebraic Operations and Polynomials

$f(1) = 1 + 2 - 5 + 1 = -1$.

$f(2) = 8 - 12 + 8 - 4 = 0$. By the factor theorem, a value of zero means $(x-2)$ divides $f(x)$ exactly.

Test $x = -1$: $-1 + 2 + 5 - 6 = 0$, so $(x+1)$ is a factor. Dividing gives $x^2 + x - 6 = (x+3)(x-2)$.

$f(3) = 27 - 18 + 3a + 6 = 15 + 3a$. Setting this to zero gives $3a = -15$, so $a = -5$.

$f(1) = 1 - 6 + 11 - 6 = 0$, so $(x-1)$ is a factor. Dividing gives $x^2 - 5x + 6 = (x-2)(x-3)$, so the roots are $1$, $2$ and $3$.

Factorising: $\dfrac{(2x-1)(x+3)}{(x-3)(x+3)}$. Cancelling $(x+3)$ leaves $\dfrac{2x-1}{x-3}$. The original denominator is zero at $x = 3$ and $x = -3$, so it is undefined there.