Prove by factor theorem that,
(iii) $(3 x - 2) is a factor of 18 x^{3} - 3 x^{2} + 6 x - 8$ .

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Solution

To prove: $(3 x - 2) is a factor of 18 x^{3} - 3 x^{2} + 6 x - 8$ .
From Factor theorem, we know that A polynomial $f (x)$ has a factor $(x - a)$ if and only if $f (a) = 0$ .
$3 x - 2 = 0 \Rightarrow x = \frac{2}{3}$
So, $a = \frac{2}{3}$
$f (a) = f (\frac{2}{3})$
$= 18 {(\frac{2}{3})}^{3} - 3 {(\frac{2}{3})}^{2} + 6 (\frac{2}{3}) - 8$
$= 18 (\frac{8}{27}) - 3 (\frac{4}{9}) + 4 - 8$
$= (\frac{16}{3}) - (\frac{4}{3}) - 4$
$= (\frac{12}{3}) - 4$
$= 0$
Hence, $(3 x - 2) is a factor of 18 x^{3} - 3 x^{2} + 6 x - 8$ .

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