If $x^{4} - 3 x^{3} + 2 x^{2} + x - 1$ is divided by $(x - 2)$ , Then remainder will be

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Solution

Given that:
$P (x) = x^{4} - 3 x^{3} + 2 x^{2} + x - 1$
WE know that if $P (x)$ be a polynomial then there exist two polynomial $Q (x) & R (x)$ such that:
$P (x) = (x - 2) Q (x) + R (x) . . . . . . . . . . . (1)$
Now
Rearrange given polynomial as:
$x^{4} - 3 x^{3} + 2 x^{2} + x - 1 = x^{4} - 2 x^{3} - x^{3} + 2 x^{2} + x - 2 + 1$
$x^{4} - 3 x^{3} + 2 x^{2} + x - 1 = x^{3} (x - 2) - x^{2} (x - 2) + 1 (x - 2) + 1$
$x^{4} - 3 x^{3} + 2 x^{2} + x - 1 = (x^{3} - x^{2} + 1) (x - 2) + 1$
$P (x) = (x^{3} - x^{2} + 10) (x - 2) + 1.............. (2)$
Compering equation $1 & 2$
$R (x) = 1$
$R e m a i n d e r i s 1 w h e n P (x) i s d i v i d e d b y x - 2$

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