Using factor theorem show that $(x - 1)$ is factor of $4 x^{3} - 6 x^{2} + 9 x - 7$ .

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Solution

According to factor theorem, when $f (x)$ is divided by $(x - 1), f (1) = 0$
$f (x) = 4 x^{3} - 6 x^{2} + 9 x - 7$
$f (1) = 4 (1)^{3} - 6 (1)^{2} + 9 (1) - 7$
$= 4 - 6 + 9 - 7 = 0$
$∵ f (1) = 0, (x - 1)$ is a factor of the given polynomial

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