Factorize $6 x^{2} + 17 x + 5$ by using the Factor Theorem.

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Solution

Let $p (x) = 6 x^{2} + 17 x + 5 = 6 (x^{2} + \frac{17}{6} + \frac{5}{6})$
The term $\frac{5}{6}$ can be written as a multiple of many terms. For example, one would be $(\pm \frac{5}{2}, \pm \frac{1}{3}), (\pm \frac{5}{6}, \pm 1), . . . \dots \dots . . .$ .
Let $p (\frac{- 1}{3}) = 0$ then by factor theorem $\frac{- 1}{3}$ is the factor of $P (x)$ then the other factor will be $\frac{- 5}{2}$ ;
Thus $6 x^{2} + 17 x + 5$ can be factorized as $(2 x + 5) (3 x + 1)$

Or,
$p (x) = 6 x^{2} + 17 x + 5 = 6 x^{2} + 15 x + 2 x + 5$
$= 2 x (3 x + 1) + 5 (3 x + 1)$
$= (2 x + 5) (3 x + 1)$

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