x^4 +x^3 −8x^2 +2x+4=0 x=1 ∨ x=2 ∨ x=2±(√2) Is this right? I have not enough time to edit my solution


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Question Number 215625 by MathematicalUser2357 last updated on 12/Jan/25

$x^{4} + x^{3} - 8 x^{2} + 2 x + 4 = 0$ $x = 1 \lor x = 2 \lor x = 2 \pm \sqrt{2}$ $Is this right ? I have not enough time to edit my solution$
Commented by MathematicalUser2357 last updated on 12/Jan/25

$why did you post late ? ? ?$
Commented by mr W last updated on 12/Jan/25

$i f y o u a r e n o t s l e e p i n g, y o u s h o u l d$ $b e a b l e t o k n o w b y y o u s e l f t h a t$ $y o u r r e s u l t i s w r o n g, s i n c e$ $1 + 2 + (2 + \sqrt{2}) + (2 - \sqrt{2}) \neq - 1$ $t h e c o r r e c t a n s w e r s h o u l d b e$ $x = 1 \lor x = 2 \lor x = - 2 \pm \sqrt{2}$ $t h e r e f o r e t h e s o m e o n e w h o i s$ $s l e e p i n g i s a c t u a l l y y o u s e l f .$
Commented by MathematicalUser2357 last updated on 12/Jan/25

$I think someone is sleeping instead of answering$
Commented by mr W last updated on 12/Jan/25

$n o b o d y m u s t a n s w e r y o u a t a l l!$ $i f y o u s e r i o u s l y n e e d h e l p, j u s t p o s t$ $y o u r q u e s t i o n s a n d w a i t . w e i r d$ $c o m m e n t s l i k e t h o s e a b o v e a r e j u s t$ $a n n o y i n g . t h e y h e l p n e i t h e r y o u s e l f$ $n o r o t h e r s .$
Commented by MathematicalUser2357 last updated on 13/Jan/25

$I failed in x = 2 \pm \sqrt{2}$
	Answered by MathematicalUser2357 last updated on 13/Jan/25

	$x^{4} + x^{3} - 8 x^{2} + 2 x + 4 = 0$ $P (x) = x^{4} + x^{3} - 8 x^{2} + 2 x + 4 \Rightarrow P (1) = 0$ $\begin{array}{cccccc} Order & Dividend / Remainder & Divisor & Quotient & Removal \\ 1 & x^{4} + x^{3} - 8 x^{2} + 2 x + 4 & x - 1 & x^{3} & x^{4} - x^{3} \\ 2 & 2 x^{3} - 8 x^{2} + 2 x + 4 & 2 x^{2} & 2 x^{3} - 2 x^{2} \\ 3 & - 6 x^{2} + 2 x + 4 & - 6 x & - 6 x^{2} + 6 x \\ 4 & - 4 x + 4 & - 4 & - 4 x + 4 \\ Total & 0 & x^{3} + 2 x^{2} - 6 x - 4 & x^{4} + x^{3} - 8 x^{2} + 2 x + 4 \end{array}$ $∴ P (x) = (x - 1) (x^{3} + 2 x^{2} - 6 x - 4) \Rightarrow x = 1$ $Q (x) = x^{3} + 2 x^{2} - 6 x - 4 \Rightarrow Q (2) = 0$ $\begin{array}{ccccc} Order & Dividend / Remainder & Divisor & Quotient & Removal \\ 1 & x^{3} + 2 x^{2} - 6 x - 4 & x - 2 & x^{2} & x^{3} - 2 x^{2} \\ 2 & 4 x^{2} - 6 x - 4 & 4 x & 4 x^{2} - 8 x \\ 3 & 2 x - 4 & 2 & 2 x - 4 \\ Total & 0 & x^{2} + 4 x + 2 & x^{3} + 2 x^{2} - 6 x - 4 \end{array}$ $∴ Q (x) = (x - 2) (x^{2} + 4 x + 2) \Rightarrow x = 2 or x = - 2 \pm \sqrt{2}$ $Welp, one mistake on mondays .$
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