x^5 +ax^4 +cx^2 +dx+e=0 let x=((rt+s)/(t+p)) . Find r,s,p such that equation gets transformed to λt^5 +Dt+E=0.


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Question Number 56641 by ajfour last updated on 20/Mar/19

$x^{5} + {ax}^{4} + {cx}^{2} + dx + e = 0$ $let x = \frac{rt + s}{t + p} . Find r, s, p such$ $that equation gets transformed$ $to λ t^{5} + Dt + E = 0 .$
Commented by ajfour last updated on 20/Mar/19

$Granted this is done, we need to$ $see if the quintic can perhaps now$ $be solved . .$ $let$ $x^{5} + kx + q = 0$ $equivalently$ $(x^{2} + fx + g) (x^{3} + {hx}^{2} + mx + n) = 0$ $\Rightarrow$ $x^{5} + (h + f) x^{4} + (m + fh + g) x^{3} +$ $(n + fm + gh) x^{2} + (fn + gm) x + gn = 0$ $\Rightarrow$ $h + f = 0$ $m + fh + g = 0$ $n + fm + gh = 0 . . . (i)$ $fn + gm = k . . . . (ii)$ $gn = q$ $from (i)$ $[\frac{q}{g} - h (h^{2} - g) + gh = 0] \times h$ $& from (ii)$ $- \frac{hq}{g} + g (h^{2} - g) = k$ $Adding$ ${3 gh}^{2} + g^{2} - h^{4} = k] \times g . . . (I) &$ ${gh}^{3} - {2 g}^{2} h = q] \times h$ $Adding$ $g^{2} h^{2} + g^{3} = gk + hq . . . (II)$ $from (I)$ $h^{2} = \frac{3 g \pm \sqrt{{9 g}^{2} - 4 (k - g^{2})}}{2}$ $\Rightarrow h^{2} = \frac{3 g}{2} \pm \frac{\sqrt{{13 g}^{2} - 4 k}}{2}$ $while from (II)$ $h = \frac{q \pm \sqrt{q^{2} - {4 g}^{3} (g^{2} - k)}}{{2 g}^{2}}$ $\Rightarrow \frac{3 g}{2} \pm \frac{\sqrt{{13 g}^{2} - 4 k}}{2} = {[\frac{q \pm \sqrt{q^{2} - {4 g}^{3} (g^{2} - k)}}{{2 g}^{2}}]}^{2}$ $+ \times ? % % / Degree 5 is degree 5 / /$
Commented by Tawa1 last updated on 20/Mar/19

$Wow, God bless you sir .$
	Answered by ajfour last updated on 20/Mar/19

	${(rt + s)}^{5} + a {(rt + s)}^{4} (t + p) + c {(rt + s)}^{2} {(t + p)}^{3}$ $+ d (rt + s) {(t + p)}^{4} + {(t + p)}^{5} = 0$ $r^{5} t^{5} + {5 r}^{4} {st}^{4} + {10 r}^{3} s^{2} t^{3} + {10 r}^{2} s^{3} t^{2} +$ $+ {5 rs}^{4} t + s^{5}$ $+ a (r^{4} t^{4} + {4 r}^{3} {st}^{3} + {6 r}^{2} s^{2} t^{2} + {4 rs}^{3} t + s^{4}) (t + p)$ $+ c (r^{2} t^{2} + 2 rst + s^{2}) (t^{3} + {3 pt}^{2} + {3 p}^{2} t + p^{3})$ $+ d (rt + s) (t^{4} + {4 pt}^{3} + {6 p}^{2} t^{2} + {4 p}^{3} t + p^{4})$ $+ e (t^{5} + {5 pt}^{4} + {10 p}^{2} t^{3} + {10 p}^{3} t^{2} + {5 p}^{4} t + p^{5})$ $= 0$ $\Rightarrow$ $(r^{5} + {ar}^{4} + {cr}^{2} + dr + e) t^{5} +$ $({5 r}^{4} s + {apr}^{4} + {4 ar}^{3} s + {3 cpr}^{2} + 2 crs +$ $4 dpr + ds + 5 ep) t^{4} +$ $({10 r}^{3} s^{2} + {4 apr}^{3} s + {6 ar}^{2} s^{2} + {3 cp}^{2} r^{2}$ $+ 6 cprs + {cs}^{2} + {6 dp}^{2} r + 4 dps + {10 ep}^{2}) t^{3}$ $({10 r}^{2} s^{3} + {6 apr}^{2} s^{2} + {4 ars}^{3} + {cp}^{3} r^{2} + {6 cp}^{2} rs$ $+ {3 cps}^{2} + {4 dp}^{3} r + {6 dp}^{2} s + {10 ep}^{3}) t^{2}$ $+ Dt + E = 0$ $If coefficients of t^{4}, t^{3}, t^{2} are to be zero;$ $then$ ${5 r}^{4} s + {apr}^{4} + {4 ar}^{3} s + {3 cpr}^{2} + 2 crs +$ $4 dpr + ds + 5 pe = 0$ $&$ ${10 r}^{3} s^{2} + {4 apr}^{3} s + {6 ar}^{2} s^{2} + {3 cp}^{2} r^{2} +$ $6 cprs + {cs}^{2} + {6 dp}^{2} r + 4 dps + {10 ep}^{2} = 0$ $&$ ${10 r}^{2} s^{3} + {6 apr}^{2} s^{2} + {4 ars}^{3} + {cp}^{3} r^{2} + {6 cp}^{2} rs$ $+ {3 cps}^{2} + {4 dp}^{3} r + {6 dp}^{2} s + {10 ep}^{3} = 0$ $. . . . .$
	Commented by Tawa1 last updated on 20/Mar/19

	$Wow, God bless you sir . Weldone$
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