Using the factor theorem factorise the following polynomial completely $3 x^{3} + 2 x^{2} - 19 x + 6$ .

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Solution

Let $f (x) = 3 x^{3} + 2 x^{2} - 19 x + 6$
For $x = 2$ , the value of $f (x)$ will be,
$= 3 (2)^{3} + 2 (2)^{2} - 19 (2) + 6$
$\Rightarrow 24 + 8 - 38 + 6 = 0$
According to the factor theorem, $(x - a)$ is a factor of $f (x)$ if $f (a) = 0$
$∴ (x - 2)$ is a factor of $f (x)$ . $[∵ f (2) = 0]$
Now, performing long division as shown in the above image, we have
$f (x) = (x - 2) (3 x^{2} + 8 x - 3)$
$\Rightarrow (x - 2) (3 x^{2} + 9 x - x - 3)$
$\Rightarrow (x - 2) [3 x (x + 3) - 1 (x + 3)]$
$\Rightarrow (x - 2) (x + 3) (3 x - 1)$

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