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

Solution $1:$

(By splitting method) : If we can find two numbers $p$ and $q$ such that $p + q = 17$ and $pq = 6 \times 5 = 30$, then we can get the factors.

So, let us look for the pairs of factors of $30$. Some are $1$ and $30$, $2$ and $15$, $3$ and $10$, $5$ and $6$. Of these pairs, $2$ and $15$ will give us $p + q = 17$.

So,  $6x^2 + 17x + 5 = 6x^2 + (2 + 15)x + 5 $

                            $= 6x^2 + 2x + 15x + 5 $

                            $= 2x(3x + 1) + 5(3x + 1) $

                             $= (3x + 1) (2x + 5)$

Solution $2:$

(Using the Factor Theorem)

$6 x^{2}+17 x+5=6\left(x^{2}+\frac{17}{6} x+\frac{5}{6}\right)=6 p(x),$ say. If $a$ and $b$ are the zeroes of $p(x),$ then $6 x^{2}+17 x+5=6(x-a)(x-b) .$ So, $a b=\frac{5}{6} \cdot$ Let us look at some possibilities for $a$ and $b$ They could be $\pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{5}{2}, \pm 1 .$ Now, $p\left(\frac{1}{2}\right)=\frac{1}{4}+\frac{17}{6}\left(\frac{1}{2}\right)+\frac{5}{6} \neq 0 .$ But $p\left(\frac{-1}{3}\right)=0 .$ So $,\left(x+\frac{1}{3}\right)$ is a factor of $p(x) .$ Similarly, by trial, you can find that $\left(x+\frac{5}{2}\right)$ is a factor of $p(x)$.

Therefore,              $6 x^{2}+17 x+5=6\left(x+\frac{1}{3}\right)\left(x+\frac{5}{2}\right)$

 $=6\left(\frac{3 x+1}{3}\right)\left(\frac{2 x+5}{2}\right)$

$=(3 x+1)(2 x+5)$

For the example above, the use of the splitting method appears more efficient. However, let us consider another example.