Factorise $6x^2 + 17x + 5$ by splitting the middle term,and by using the Factor Theorem.

(N/A) Solution $1:$ (By splitting the middle term method)
We need to find two numbers $p$ and $q$ such that $p + q = 17$ and $pq = 6 \times 5 = 30$.
The factors of $30$ are $(1, 30), (2, 15), (3, 10), (5, 6)$.
Among these,$2 + 15 = 17$.
So,$6x^2 + 17x + 5 = 6x^2 + 2x + 15x + 5$
$= 2x(3x + 1) + 5(3x + 1)$
$= (3x + 1)(2x + 5)$.
Solution $2:$ (Using the Factor Theorem)
Let $p(x) = 6x^2 + 17x + 5 = 6(x^2 + \frac{17}{6}x + \frac{5}{6})$.
Let $a$ and $b$ be the zeroes of the quadratic expression. Then $ab = \frac{5}{6}$.
Possible rational roots are $\pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{5}{2}, \pm 1$.
Testing $p(-\frac{1}{3}) = 6(-\frac{1}{3})^2 + 17(-\frac{1}{3}) + 5 = 6(\frac{1}{9}) - \frac{17}{3} + 5 = \frac{2}{3} - \frac{17}{3} + \frac{15}{3} = 0$.
Thus,$(x + \frac{1}{3})$ is a factor.
Testing $p(-\frac{5}{2}) = 6(-\frac{5}{2})^2 + 17(-\frac{5}{2}) + 5 = 6(\frac{25}{4}) - \frac{85}{2} + 5 = \frac{75}{2} - \frac{85}{2} + \frac{10}{2} = 0$.
Thus,$(x + \frac{5}{2})$ is a factor.
Therefore,$6x^2 + 17x + 5 = 6(x + \frac{1}{3})(x + \frac{5}{2}) = 6(\frac{3x+1}{3})(\frac{2x+5}{2}) = (3x + 1)(2x + 5)$.