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

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} \dots (i)

2 q = (B + p^{2}) - \frac{C}{p} \dots (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 \dots (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 \dots

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 .

Facebook Tweet Pin