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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 4th degree and solve. Or probably a polynomial of nth power to smaller power.
$$\mathrm{Please}\:\mathrm{is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{way}\:\mathrm{to}\:\mathrm{reduce}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{of}\:\:\mathrm{4th}\:\mathrm{degree} \\ $$$$\mathrm{and}\:\mathrm{solve}.\:\:\mathrm{Or}\:\mathrm{probably}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{of}\:\:\:\mathrm{nth}\:\mathrm{power}\:\mathrm{to}\:\mathrm{smaller} \\ $$$$\mathrm{power}.\: \\ $$
Answered by ajfour last updated on 17/Mar/19
x^4 +ax^3 +bx^2 +cx+d=0 let x=t−(a/4) t^4 −at^3 +(3/8)a^2 t^2 −(a^3 /(16))t+(a^4 /(256)) +at^3 −((3a^2 )/4)t^2 +((3a^3 )/(16))t−(a^4 /(64)) +bt^2 −((ab)/2)t+((a^2 b)/(16))+ct−((ac)/4)+d=0 ⇒ t^4 +(b−((3a^2 )/8))t^2 +((a^3 /8)−((ab)/2)+c)t +(((3a^4 )/(256))+((a^2 b)/(16))−((ac)/4)+d)=0 Say we have now t^4 +Bt^2 +Ct+D = (t^2 +pt+q)(t^2 −pt+r)=0 ⇒ t^4 +(r−p^2 +q)t^2 +p(r−q)t+qr=0 ⇒ q+r=B+p^2 r−q=C/p qr=D ⇒ 2r=(B+p^2 )+(C/p) ...(i) 2q=(B+p^2 )−(C/p) ...(ii) ⇒ (B+p^2 )^2 −(C^( 2) /p^2 )=4D let p^2 =z ⇒ z(B+z)^2 −4Dz−C^( 2) =0 ⇒ z^3 +2Bz^2 +(B^2 −4D)z−C^( 2) =0 Find z=p^2 from above eq. Then using (i)&(ii) obtain q and r. Now we have (t^2 +pt+q)(t^2 −pt+r)=0 ⇒ t=−(p/2)±(√((p^2 /4)−q)) or t= (p/2)±(√((p^2 /4)−r)) & x=t−(a/4) .
$$ \\ $$$$ \\ $$$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$${let}\:\:{x}={t}−\frac{{a}}{\mathrm{4}} \\ $$$${t}^{\mathrm{4}} −{at}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{8}}{a}^{\mathrm{2}} {t}^{\mathrm{2}} −\frac{{a}^{\mathrm{3}} }{\mathrm{16}}{t}+\frac{{a}^{\mathrm{4}} }{\mathrm{256}} \\ $$$$+{at}^{\mathrm{3}} −\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}}{t}^{\mathrm{2}} +\frac{\mathrm{3}{a}^{\mathrm{3}} }{\mathrm{16}}{t}−\frac{{a}^{\mathrm{4}} }{\mathrm{64}} \\ $$$$+{bt}^{\mathrm{2}} −\frac{{ab}}{\mathrm{2}}{t}+\frac{{a}^{\mathrm{2}} {b}}{\mathrm{16}}+{ct}−\frac{{ac}}{\mathrm{4}}+{d}=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{4}} +\left({b}−\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{8}}\right){t}^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{3}} }{\mathrm{8}}−\frac{{ab}}{\mathrm{2}}+{c}\right){t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{\mathrm{3}{a}^{\mathrm{4}} }{\mathrm{256}}+\frac{{a}^{\mathrm{2}} {b}}{\mathrm{16}}−\frac{{ac}}{\mathrm{4}}+{d}\right)=\mathrm{0} \\ $$$${Say}\:{we}\:{have}\:{now} \\ $$$$\:\:\:\:{t}^{\mathrm{4}} +{Bt}^{\mathrm{2}} +{Ct}+{D} \\ $$$$\:\:\:=\:\left({t}^{\mathrm{2}} +{pt}+{q}\right)\left({t}^{\mathrm{2}} −{pt}+{r}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{4}} +\left({r}−{p}^{\mathrm{2}} +{q}\right){t}^{\mathrm{2}} +{p}\left({r}−{q}\right){t}+{qr}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{q}+{r}={B}+{p}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:{r}−{q}={C}/{p} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{qr}={D} \\ $$$$\Rightarrow\:\:\:\mathrm{2}{r}=\left({B}+{p}^{\mathrm{2}} \right)+\frac{{C}}{{p}}\:\:\:\:\:\:\:…\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{q}=\left({B}+{p}^{\mathrm{2}} \right)−\frac{{C}}{{p}}\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\:\:\:\:\:\left({B}+{p}^{\mathrm{2}} \right)^{\mathrm{2}} −\frac{{C}^{\:\mathrm{2}} }{{p}^{\mathrm{2}} }=\mathrm{4}{D} \\ $$$$\:\:\:\:{let}\:\:{p}^{\mathrm{2}} ={z} \\ $$$$\Rightarrow\:\:\:{z}\left({B}+{z}\right)^{\mathrm{2}} −\mathrm{4}{Dz}−{C}^{\:\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{z}^{\mathrm{3}} +\mathrm{2}{Bz}^{\mathrm{2}} +\left({B}^{\mathrm{2}} −\mathrm{4}{D}\right){z}−{C}^{\:\mathrm{2}} =\mathrm{0} \\ $$$${Find}\:{z}={p}^{\mathrm{2}} \:{from}\:{above}\:{eq}. \\ $$$${Then}\:{using}\:\left({i}\right)\&\left({ii}\right)\:{obtain} \\ $$$${q}\:{and}\:{r}. \\ $$$${Now}\:{we}\:{have} \\ $$$$\:\:\left({t}^{\mathrm{2}} +{pt}+{q}\right)\left({t}^{\mathrm{2}} −{pt}+{r}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}=−\frac{{p}}{\mathrm{2}}\pm\sqrt{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}}\:\:\:{or} \\ $$$$\:\:\:\:\:\:\:\:{t}=\:\frac{{p}}{\mathrm{2}}\pm\sqrt{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{r}}\:\: \\ $$$$\&\:\:\:\:{x}={t}−\frac{{a}}{\mathrm{4}}\:. \\ $$
Commented by Tawa1 last updated on 17/Mar/19
God bless you sir, i look forward for the success sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\:\mathrm{i}\:\mathrm{look}\:\mathrm{forward}\:\mathrm{for}\:\mathrm{the}\:\mathrm{success}\:\mathrm{sir} \\ $$
Commented by ajfour last updated on 17/Mar/19
Or after t^4 +Bt^2 +Ct+D=0 we let x=((pt+q)/(t+1)) then (pt+q)^4 +B(pt+q)^2 (t+1)^2 + C(pt+q)(t+1)^3 +D(t+1)^4 =0 ⇒ p^4 t^4 +4p^3 qt^3 +6p^2 q^2 t^2 +4pq^3 t+q^4 + B(p^2 t^2 +2pqt+q^2 )(t^2 +2t+1)+ C(pt+q)(t^3 +3t^2 +3t+1)+ D(t^4 +4t^3 +6t^2 +4t+1)=0 ⇒ __________________________ (p^4 +Bp^2 +Cp+D)t^4 + (4p^3 q+2Bp^2 +2Bpq+3Cp+Cq+4D)t^3 +(6p^2 q^2 +Bp^2 +4Bpq+Bq^2 +3Cp+3Cq+6D)t^2 +(4pq^3 +2Bpq+2Bq^2 +Cp+3Cq+4D)t +(q^4 +Bq^2 +Cq+D)=0 __________________________ & if we let coefficients of t^3 & t to be zero we obtain a quadratic in t^2 . 4p^3 q+2Bp^2 +2Bpq+3Cp+Cq+4D=0 4pq^3 +2Bpq+2Bq^2 +Cp+3Cq+4D=0 subtracting we get 4pq(p+q)+2B(p+q)+2C=0 ..(I) And adding we get (4pq+2B)(p^2 +q^2 )+4Bpq+ 4C(p+q)+8D=0 ...(II) Solving this pair of eqs. (I)&(II), we′d have reduced a biquadratic to a quadratic. rewriting eq. (II) (4pq+2B)[(p+q)^2 −2pq]+4Bpq +4C(p+q)+8D=0 ⇒ 4C(p+q)+8D=−(4pq+2B)(p+q)^2 using (I) herein ⇒ 4C(p+q)+8D=2(p+q)[C+B(p+q)] ⇒ B(p+q)^2 −C(p+q)−4D=0 ⇒ p+q = (C/2)±(√((C^( 2) /4)+4D)) = k pq = −(B/2)−(C/(2(p+q))) = −(B/2)−(C/(2k)) = l p, q are roots of z^2 −kz+l=0 p, q = (k/2)±(√((k^2 /4)−l)) Now we have (p^4 +Bp^2 +Cp+D)t^4 +(6p^2 q^2 +Bp^2 +4Bpq+Bq^2 +3Cp+3Cq+6D)t^2 +(q^4 +Bq^2 +Cq+D)=0 which is a quadratic in t^2 , and can be solved. And x=((pt+q)/(t+1)). ■
$${Or}\:{after}\: \\ $$$$\:\:\:\:\:{t}^{\mathrm{4}} +{Bt}^{\mathrm{2}} +{Ct}+{D}=\mathrm{0} \\ $$$${we}\:{let}\:\:\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}}\:\:\:{then} \\ $$$$\:\:\left({pt}+{q}\right)^{\mathrm{4}} +{B}\left({pt}+{q}\right)^{\mathrm{2}} \left({t}+\mathrm{1}\right)^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:{C}\left({pt}+{q}\right)\left({t}+\mathrm{1}\right)^{\mathrm{3}} +{D}\left({t}+\mathrm{1}\right)^{\mathrm{4}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\:\:\:{p}^{\mathrm{4}} {t}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{3}} {qt}^{\mathrm{3}} +\mathrm{6}{p}^{\mathrm{2}} {q}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{4}{pq}^{\mathrm{3}} {t}+{q}^{\mathrm{4}} \\ $$$$+\:{B}\left({p}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{2}{pqt}+{q}^{\mathrm{2}} \right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}\right)+ \\ $$$$\:\:{C}\left({pt}+{q}\right)\left({t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{1}\right)+ \\ $$$$\:\:{D}\left({t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{3}} +\mathrm{6}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\:\left({p}^{\mathrm{4}} +{Bp}^{\mathrm{2}} +{Cp}+{D}\right){t}^{\mathrm{4}} + \\ $$$$\:\:\:\:\left(\mathrm{4}{p}^{\mathrm{3}} {q}+\mathrm{2}{Bp}^{\mathrm{2}} +\mathrm{2}{Bpq}+\mathrm{3}{Cp}+{Cq}+\mathrm{4}{D}\right){t}^{\mathrm{3}} \\ $$$$+\left(\mathrm{6}{p}^{\mathrm{2}} {q}^{\mathrm{2}} +{Bp}^{\mathrm{2}} +\mathrm{4}{Bpq}+{Bq}^{\mathrm{2}} +\mathrm{3}{Cp}+\mathrm{3}{Cq}+\mathrm{6}{D}\right){t}^{\mathrm{2}} \\ $$$$+\left(\mathrm{4}{pq}^{\mathrm{3}} +\mathrm{2}{Bpq}+\mathrm{2}{Bq}^{\mathrm{2}} +{Cp}+\mathrm{3}{Cq}+\mathrm{4}{D}\right){t} \\ $$$$+\left({q}^{\mathrm{4}} +{Bq}^{\mathrm{2}} +{Cq}+{D}\right)=\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\&\:{if}\:{we}\:{let}\:{coefficients}\:{of}\:{t}^{\mathrm{3}} \:\&\:{t} \\ $$$${to}\:{be}\:{zero}\:{we}\:{obtain}\:{a}\:{quadratic} \\ $$$${in}\:{t}^{\mathrm{2}} . \\ $$$$\:\mathrm{4}{p}^{\mathrm{3}} {q}+\mathrm{2}{Bp}^{\mathrm{2}} +\mathrm{2}{Bpq}+\mathrm{3}{Cp}+{Cq}+\mathrm{4}{D}=\mathrm{0} \\ $$$$\:\mathrm{4}{pq}^{\mathrm{3}} +\mathrm{2}{Bpq}+\mathrm{2}{Bq}^{\mathrm{2}} +{Cp}+\mathrm{3}{Cq}+\mathrm{4}{D}=\mathrm{0} \\ $$$${subtracting}\:{we}\:{get} \\ $$$$\:\mathrm{4}{pq}\left({p}+{q}\right)+\mathrm{2}{B}\left({p}+{q}\right)+\mathrm{2}{C}=\mathrm{0}\:\:..\left({I}\right) \\ $$$${And}\:{adding}\:{we}\:{get} \\ $$$$\left(\mathrm{4}{pq}+\mathrm{2}{B}\right)\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)+\mathrm{4}{Bpq}+ \\ $$$$\:\:\:\:\:\:\mathrm{4}{C}\left({p}+{q}\right)+\mathrm{8}{D}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({II}\right) \\ $$$${Solving}\:{this}\:{pair} \\ $$$${of}\:{eqs}.\:\left({I}\right)\&\left({II}\right), \\ $$$$\:{we}'{d}\:{have}\:{reduced}\:{a}\: \\ $$$${biquadratic}\:{to}\:{a}\:{quadratic}. \\ $$$${rewriting}\:{eq}.\:\left({II}\right) \\ $$$$\left(\mathrm{4}{pq}+\mathrm{2}{B}\right)\left[\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{2}{pq}\right]+\mathrm{4}{Bpq} \\ $$$$\:\:\:\:\:\:\:+\mathrm{4}{C}\left({p}+{q}\right)+\mathrm{8}{D}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{4}{C}\left({p}+{q}\right)+\mathrm{8}{D}=−\left(\mathrm{4}{pq}+\mathrm{2}{B}\right)\left({p}+{q}\right)^{\mathrm{2}} \\ $$$${using}\:\left({I}\right)\:{herein} \\ $$$$\Rightarrow\:\mathrm{4}{C}\left({p}+{q}\right)+\mathrm{8}{D}=\mathrm{2}\left({p}+{q}\right)\left[{C}+{B}\left({p}+{q}\right)\right] \\ $$$$\Rightarrow\:{B}\left({p}+{q}\right)^{\mathrm{2}} −{C}\left({p}+{q}\right)−\mathrm{4}{D}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{p}+{q}\:=\:\frac{{C}}{\mathrm{2}}\pm\sqrt{\frac{{C}^{\:\mathrm{2}} }{\mathrm{4}}+\mathrm{4}{D}}\:\:=\:{k} \\ $$$$\:\:\:\:{pq}\:=\:−\frac{{B}}{\mathrm{2}}−\frac{{C}}{\mathrm{2}\left({p}+{q}\right)}\: \\ $$$$\:\:\:\:\:\:\:\:\:=\:−\frac{{B}}{\mathrm{2}}−\frac{{C}}{\mathrm{2}{k}}\:=\:{l} \\ $$$$\:{p},\:{q}\:{are}\:{roots}\:{of} \\ $$$$\:\:\:\:\:\:\:\:\:{z}^{\mathrm{2}} −{kz}+{l}=\mathrm{0} \\ $$$${p},\:{q}\:=\:\frac{{k}}{\mathrm{2}}\pm\sqrt{\frac{{k}^{\mathrm{2}} }{\mathrm{4}}−{l}} \\ $$$${Now}\:{we}\:{have} \\ $$$$\:\:\:\left({p}^{\mathrm{4}} +{Bp}^{\mathrm{2}} +{Cp}+{D}\right){t}^{\mathrm{4}} \\ $$$$+\left(\mathrm{6}{p}^{\mathrm{2}} {q}^{\mathrm{2}} +{Bp}^{\mathrm{2}} +\mathrm{4}{Bpq}+{Bq}^{\mathrm{2}} +\mathrm{3}{Cp}+\mathrm{3}{Cq}+\mathrm{6}{D}\right){t}^{\mathrm{2}} \\ $$$$+\left({q}^{\mathrm{4}} +{Bq}^{\mathrm{2}} +{Cq}+{D}\right)=\mathrm{0} \\ $$$$\:\:\:{which}\:{is}\:{a}\:{quadratic}\:{in}\:{t}^{\mathrm{2}} , \\ $$$${and}\:{can}\:{be}\:{solved}. \\ $$$${And}\:\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}}.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare \\ $$
Commented by Tawa1 last updated on 17/Mar/19
Wow, God bless you sir. I appreciate ...
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}\:… \\ $$
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.
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{if}\:\mathrm{the}\:\mathrm{power}\:\mathrm{is}\:\:\mathrm{5}\:\mathrm{or}\:\mathrm{higher}.\:\mathrm{can}\:\mathrm{we}\:\mathrm{still}\:\mathrm{deduce}. \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\ $$

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