(x^4 −3x^2 +2x+1)/(x−1)


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Question Number 60854 by Askash last updated on 26/May/19

$(x^{4} - 3 x^{2} + 2 x + 1) / (x - 1)$
Commented by Askash last updated on 26/May/19

$i w a n t t h e b e s t m e t h o d .$
Commented by maxmathsup by imad last updated on 26/May/19

$l e t p (x) = x^{4} - 3 x^{2} + 2 x + 1 t h e b e s t m e t h o d i s t o d i v i d e p (x) b y x - 1$ $(u s i n g p p o p e r a t i o n) l e t u s e a n o t h e r m e t h o d w e h a v e d e g (x - 1) = 1 \Rightarrow$ $p (x) = (x - 1) (x^{3} + {a x}^{2} + b x + c) + d d = p (1) = 1 - 3 + 2 + 1 = 1 \Rightarrow$ $p (x) = (x - 1) (x^{3} + {a x}^{2} + b x + c) + 1 \Rightarrow$ $x^{4} + {a x}^{3} + {b x}^{2} + c x - x^{3} - {a x}^{2} - b x - c + 1 = x^{4} - 3 x^{2} + 2 x + 1 \Rightarrow$ $(a - 1) x^{3} + (b - a) x^{2} + (c - b) x - c = - 3 x^{2} + 2 x \Rightarrow$ $a - 1 = 0 a n d b - a = - 3 a n d c - b = 2 \Rightarrow a = 1, b = - 2, c = 0 \Rightarrow$ $Q (x) = x^{3} + x^{2} - 2 x a n d R = 1$
Commented by Tawa1 last updated on 26/May/19

$let x - 1 = 0, \Rightarrow x = 1$ $So, by remainder theorem$ $f (x) = x^{4} - {3 x}^{2} + 2 x + 1$ $f (1) = 1^{4} - 3 {(1)}^{2} + 2 (1) + 1$ $f (1) = 1 - 3 + 2 + 1$ $f (1) = 1$ $So, the remainder is 1 .$
Commented by Askash last updated on 26/May/19

$a n y e a s y t o f i n d r e m a i n d e r$
	Answered by ajfour last updated on 26/May/19
	$let x−1=t ⇒ x=t+1 ,then {(t+1)^4 −3(t+1)^2 +2(t+1)+1}/t =(t^4 +4t^3 +6t^2 +4t+1−3t^2 −6t−3 +2t+2+1)/t =t^3 +4t^2 +3t+(1/t) = (x−1){(x−1)^2 +4(x−1)+3}+(1/(x−1)) =x(x−1)(x+2)+(1/(x−1)) .$
	$l e t x - 1 = t \Rightarrow x = t + 1, t h e n$ ${{(t + 1)}^{4} - 3 {(t + 1)}^{2} + 2 (t + 1) + 1} / t$ $= (t^{4} + 4 t^{3} + 6 t^{2} + 4 t + 1 - 3 t^{2} - 6 t - 3$ $+ 2 t + 2 + 1) / t$ $= t^{3} + 4 t^{2} + 3 t + \frac{1}{t}$ $= (x - 1) {{(x - 1)}^{2} + 4 (x - 1) + 3} + \frac{1}{x - 1}$ $= x (x - 1) (x + 2) + \frac{1}{x - 1} .$
	Answered by Rasheed.Sindhi last updated on 26/May/19

	$S y n t h e t i c D i v i s i o n i s p e r h a p s t h e$ $e a s i e s t w a y w h e n d i v i s o r i s l i n e a r$ $a s i n t h i s c a s e .$ $P (x) = x^{4} + 0 x^{3} - 3 x^{2} + 2 x + 1$ $D (x) = x - 1$ $Q (x) : Q u o t i e n t \int$ $R : R e m a i n d e r$ $(\begin{matrix} 1) & 1 & 0 & - 3 & 2 & 1 \\ 1 & 1 & - 2 & 0 \\ - & - & - & - & - \\ 1 & 1 & - 2 & 0 & [1 \end{matrix})$ $Q (x) = x^{3} + x^{2} - 2 x, R = 1$
	Commented by malwaan last updated on 26/May/19

	$y e s m r R a s h e e d$ $f a s t a n d e a s y$ $t h a n k y o u$
	Commented by Rasheed.Sindhi last updated on 27/May/19

	${Y o u}^{'} r e w e l c o m e s i r!$
	Commented by Rasheed.Sindhi last updated on 27/May/19

	$T h i s i s a l s o a n e a s y w a y t o e v a l u a t e$ $p o l y n o m i a l P (x) a t x = a i - e P (a) .$
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