Find the remainder when $4 x^{3} - 5 x^{2} + 6 x - 2$ is divided by $x - 1$ .

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Solution

According to remainder theorem, when $f (x)$ is divided by $(x - 1)$ , Remainder $= f (1)$
$f (x) = 4 x^{3} - 5 x^{2} + 6 x - 2$
$f (1) = 4 (1)^{3} - 5 (1)^{2} + 6 (1) - 2$
$= 4 - 5 + 6 - 2 = 10 - 7 = 3$
$∴$ Remanider $= 3$

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