Please is there any way to reduce a polynomial of 4th degree and solve. Or probably a polynomial of nth power to smaller power.


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Question Number 56479 by Tawa1 last updated on 17/Mar/19

$Please is there any way to reduce a polynomial of 4 th degree$ $and solve . Or probably a polynomial of nth power to smaller$ $power .$
	Answered by ajfour last updated on 17/Mar/19

	$x^{4} + {a x}^{3} + {b x}^{2} + c x + d = 0$ $l e t x = t - \frac{a}{4}$ $t^{4} - {a t}^{3} + \frac{3}{8} a^{2} t^{2} - \frac{a^{3}}{16} t + \frac{a^{4}}{256}$ $+ {a t}^{3} - \frac{3 a^{2}}{4} t^{2} + \frac{3 a^{3}}{16} t - \frac{a^{4}}{64}$ $+ {b t}^{2} - \frac{a b}{2} t + \frac{a^{2} b}{16} + c t - \frac{a c}{4} + d = 0$ $\Rightarrow t^{4} + (b - \frac{3 a^{2}}{8}) t^{2} + (\frac{a^{3}}{8} - \frac{a b}{2} + c) t$ $+ (\frac{3 a^{4}}{256} + \frac{a^{2} b}{16} - \frac{a c}{4} + d) = 0$ $S a y w e h a v e n o w$ $t^{4} + {B t}^{2} + C t + D$ $= (t^{2} + p t + q) (t^{2} - p t + r) = 0$ $\Rightarrow t^{4} + (r - p^{2} + q) t^{2} + p (r - q) t + q r = 0$ $\Rightarrow q + r = B + p^{2}$ $r - q = C / p$ $q r = D$ $\Rightarrow 2 r = (B + p^{2}) + \frac{C}{p} . . . (i)$ $2 q = (B + p^{2}) - \frac{C}{p} . . . (i i)$ $\Rightarrow {(B + p^{2})}^{2} - \frac{C^{2}}{p^{2}} = 4 D$ $l e t p^{2} = z$ $\Rightarrow z {(B + z)}^{2} - 4 D z - C^{2} = 0$ $\Rightarrow z^{3} + 2 {B z}^{2} + (B^{2} - 4 D) z - C^{2} = 0$ $F i n d z = p^{2} f r o m a b o v e e q .$ $T h e n u s i n g (i) & (i i) o b t a i n$ $q a n d r .$ $N o w w e h a v e$ $(t^{2} + p t + q) (t^{2} - p t + r) = 0$ $\Rightarrow t = - \frac{p}{2} \pm \sqrt{\frac{p^{2}}{4} - q} o r$ $t = \frac{p}{2} \pm \sqrt{\frac{p^{2}}{4} - r}$ $& x = t - \frac{a}{4} .$
	Commented by Tawa1 last updated on 17/Mar/19

	$God bless you sir, i look forward for the success sir$
	Commented by ajfour last updated on 17/Mar/19

	$O r a f t e r$ $t^{4} + {B t}^{2} + C t + D = 0$ $w e l e t x = \frac{p t + q}{t + 1} t h e n$ ${(p t + q)}^{4} + B {(p t + q)}^{2} {(t + 1)}^{2} +$ $C (p t + q) {(t + 1)}^{3} + D {(t + 1)}^{4} = 0$ $\Rightarrow$ $p^{4} t^{4} + 4 p^{3} {q t}^{3} + 6 p^{2} q^{2} t^{2} + 4 {p q}^{3} t + q^{4}$ $+ B (p^{2} t^{2} + 2 p q t + q^{2}) (t^{2} + 2 t + 1) +$ $C (p t + q) (t^{3} + 3 t^{2} + 3 t + 1) +$ $D (t^{4} + 4 t^{3} + 6 t^{2} + 4 t + 1) = 0$ $\Rightarrow$ $__________________________$ $(p^{4} + {B p}^{2} + C p + D) t^{4} +$ $(4 p^{3} q + 2 {B p}^{2} + 2 B p q + 3 C p + C q + 4 D) t^{3}$ $+ (6 p^{2} q^{2} + {B p}^{2} + 4 B p q + {B q}^{2} + 3 C p + 3 C q + 6 D) t^{2}$ $+ (4 {p q}^{3} + 2 B p q + 2 {B q}^{2} + C p + 3 C q + 4 D) t$ $+ (q^{4} + {B q}^{2} + C q + D) = 0$ $__________________________$ $& i f w e l e t c o e f f i c i e n t s o f t^{3} & t$ $t o b e z e r o w e o b t a i n a q u a d r a t i c$ $i n t^{2} .$ $4 p^{3} q + 2 {B p}^{2} + 2 B p q + 3 C p + C q + 4 D = 0$ $4 {p q}^{3} + 2 B p q + 2 {B q}^{2} + C p + 3 C q + 4 D = 0$ $s u b t r a c t i n g w e g e t$ $4 p q (p + q) + 2 B (p + q) + 2 C = 0 . . (I)$ $A n d a d d i n g w e g e t$ $(4 p q + 2 B) (p^{2} + q^{2}) + 4 B p q +$ $4 C (p + q) + 8 D = 0 . . . (I I)$ $S o l v i n g t h i s p a i r$ $o f e q s . (I) & (I I),$ ${w e}^{'} d h a v e r e d u c e d a$ $b i q u a d r a t i c t o a q u a d r a t i c .$ $r e w r i t i n g e q . (I I)$ $(4 p q + 2 B) [{(p + q)}^{2} - 2 p q] + 4 B p q$ $+ 4 C (p + q) + 8 D = 0$ $\Rightarrow 4 C (p + q) + 8 D = - (4 p q + 2 B) {(p + q)}^{2}$ $u s i n g (I) h e r e i n$ $\Rightarrow 4 C (p + q) + 8 D = 2 (p + q) [C + B (p + q)]$ $\Rightarrow B {(p + q)}^{2} - C (p + q) - 4 D = 0$ $\Rightarrow p + q = \frac{C}{2} \pm \sqrt{\frac{C^{2}}{4} + 4 D} = k$ $p q = - \frac{B}{2} - \frac{C}{2 (p + q)}$ $= - \frac{B}{2} - \frac{C}{2 k} = l$ $p, q a r e r o o t s o f$ $z^{2} - k z + l = 0$ $p, q = \frac{k}{2} \pm \sqrt{\frac{k^{2}}{4} - l}$ $N o w w e h a v e$ $(p^{4} + {B p}^{2} + C p + D) t^{4}$ $+ (6 p^{2} q^{2} + {B p}^{2} + 4 B p q + {B q}^{2} + 3 C p + 3 C q + 6 D) t^{2}$ $+ (q^{4} + {B q}^{2} + C q + D) = 0$ $w h i c h i s a q u a d r a t i c i n t^{2},$ $a n d c a n b e s o l v e d .$ $A n d x = \frac{p t + q}{t + 1} . ◼$
	Commented by Tawa1 last updated on 17/Mar/19

	$Wow, God bless you sir . I appreciate . . .$
	Commented by Tawa1 last updated on 17/Mar/19

	$Sir, please if the power is 5 or higher . can we still deduce .$ $Thanks for your time sir .$
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