# Thread: power of 5 and above

1. ## power of 5 and above

Why equations with power of 5 and above don't have a way to slove them?

x^5 = 32
x = ?

3. ## Re: power of 5 and above

If you mean "why do polynomial equations of degree 5 and higher not have a general method that applies to all such equations" it was proved that there exist such polynomials that have solutions that cannot be written in terms of roots.

See the "Abel-Ruffini theorem": https://en.wikipedia.org/wiki/Abel%E...uffini_theorem

4. ## Re: power of 5 and above

thinking out loud:
if (x-a)(x-b)(x-c)(x-d)(x-e) = 0
then at least one of a,b,c,d,e = x

Halls, should I say: stinking out loud?!

5. ## Re: power of 5 and above

Whatever turns you on.

6. ## Re: power of 5 and above

Originally Posted by DenisB
thinking out loud:
if (x-a)(x-b)(x-c)(x-d)(x-e) = 0
then at least one of a,b,c,d,e = x

Halls, should I say: stinking out loud?!
(1)How I know that:
(a - x)(b - x)...(z - x)
have a power greater than 3?
(2)Why the equation is polynomial equation?
(3)How I know that equation that write (not like in (1) question) is polynomial that isn't seen polynomial at first sight?

7. ## Re: power of 5 and above

and post in proper English...else we can't help...

8. ## Re: power of 5 and above

Originally Posted by yyakob
(1)How I know that:
(a - x)(b - x)...(z - x)
have a power greater than 3?
(2)Why the equation is polynomial equation?
(3)How I know that equation that write (not like in (1) question) is polynomial that isn't seen polynomial at first sight?
(1) and (3): Multiply it out:

$$(x-a)(x-b)(x-c)(x-d)(x-e) = x^5-(a+b+c+d+e)x^4+(ab+ac+ad+ae+bc+bd+be+cd+ce+de)x^3-(abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde)x^2+(abcd +abce+bcde)x-abcde$$

(2) Any equation with a polynomial on one side of the equation and a constant (or another polynomial) on the other is called a polynomial equation by definition. There is no reason for it other than that is its definition.

9. ## Re: power of 5 and above

Originally Posted by yyakob
(1)How I know that:
(a - x)(b - x)...(z - x)
Is this supposed to be a trick question!?

(a - x)(b - x)...(z - x) =

(a - x)(b - x)...(w - x)(x - x)(y - x)(z - x) =

(a - x)(b - x)...(w - x)(0)(y - x)(z - x) =

0