please help me finding the roots of x^5 +5x^4 +20x^3 +60x^2 +120x+120=0?


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Question Number 141814 by gsk2684 last updated on 12/Jul/21

$p l e a s e h e l p m e f i n d i n g t h e r o o t s o f$ $x^{5} + 5 x^{4} + 20 x^{3} + 60 x^{2} + 120 x + 120 = 0 ?$
Commented by MJS_new last updated on 23/May/21

$you {don}^{'} t believe me ?$
Commented by gsk2684 last updated on 23/May/21

$D e a r M J S, I h a v e r e g i s t e r e d$ $n e w l y o n t h i s p l a t f o r m . I h a d b e l i e f$ $o n y o u . p l e a s e h e l p m e i n s o l v i n g$ $t h e p r o b l e m t o f i n d t h e n a t u r e o f$ $r o o t s a n d t r a c e o f t h e c u r v e .$
	Answered by MJS_new last updated on 23/May/21
	$f(x)=x^5 +5x^4 +20x^3 +60x^2 +120x+120 f ′(x)=5x^4 +20x^3 +60x^2 +120x+120 f ′′(x)=20x^3 +60x^2 +120x+120 f ′′(x)=0 ⇒ x≈−1.59607 f ′(≈−1.59607)≈32.4474 ⇒ f ′(x)>0 ∀x∈R ⇒ ⇒ f(x) is increasing ∀x∈R ⇒ it has exactly one real solution first we try all factors of the constant: ±{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 30, 40, 60, 120} ⇒ none of these solve the equation there′s no solution formula for polynomes of degree 5 or higher ⇒ we must approximate f(0)=120∧f(x) increasing ⇒ search <0 f(−1)=44 f(−2)=8 f(−3)=−78 f(−2.1)=3.77... f(−2.2)=−.968... f(−2.19)=−.465... f(−2.18)=.0299... ... ⇒ x≈−2.18060712404$
	$f (x) = x^{5} + 5 x^{4} + 20 x^{3} + 60 x^{2} + 120 x + 120$ $f^{'} (x) = 5 x^{4} + 20 x^{3} + 60 x^{2} + 120 x + 120$ $f^{″} (x) = 20 x^{3} + 60 x^{2} + 120 x + 120$ $f^{″} (x) = 0 \Rightarrow x \approx - 1.59607$ $f^{'} (\approx - 1.59607) \approx 32.4474 \Rightarrow f^{'} (x) > 0 \forall x \in R \Rightarrow$ $\Rightarrow f (x) is increasing \forall x \in R \Rightarrow it has exactly$ $one real solution$ $first we try all factors of the constant :$ $\pm {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 30, 40, 60, 120}$ $\Rightarrow none of these solve the equation$ ${there}^{'} s no solution formula for polynomes$ $of degree 5 or higher \Rightarrow we must approximate$ $f (0) = 120 \land f (x) increasing \Rightarrow search < 0$ $f (- 1) = 44$ $f (- 2) = 8$ $f (- 3) = - 78$ $f (- 2.1) = 3.77 . . .$ $f (- 2.2) = - . 968 . . .$ $f (- 2.19) = - . 465 . . .$ $f (- 2.18) = . 0299 . . .$ $. . .$ $\Rightarrow x \approx - 2.18060712404$
	Commented by gsk2684 last updated on 24/May/21

	$t h a n k y o u v e r y m u c h$
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