# How do I solve this equation algebraically 12x^50 + 89x^37 + 237x^21 + 122x^18 + 11x^14 - 342x^11 + 55x^9 - (12x)^7 +14x^5 + 37x^2 + 527 = 0?

### 4 Answers

- llafferLv 78 months ago
The only way I can think of to do this is to use Newton's method.

- Log in to reply to the answers

- nbsaleLv 68 months ago
You can't. There's no simple expression for the zeroes. It's possible to express the roots of polynomials up to degree 4 using radicals. Some of those roots may have very complicated expressions, so they aren't necessarily very useful. In general, for degree 5 and higher, you have to use numerical techniques.

- Log in to reply to the answers

- 8 months ago
You don't. Your best hope is to use the rational root theorem and hope that one of the possibilities works

Find divisors of 12 and 527

12 => -12 , -6 , -4 , -3 , -2 , -1 , 1 , 2 , 3 , 4 , 6 , 12

527 => -527 , -31 , -17 , -1 , 1 , 17 , 31 , 527

+/- 527/12 , +/- 527/6 , +/- 527/4 , +/- 527/3 , +/- 527/2 , +/- 527

+/- 31/12 , +/- 31/6 , +/- 31/4 , +/- 31/3 , +/- 31/2 , +/- 31

+/- 17/12 , +/- 17/6 , +/- 17/4 , +/- 17/3 , +/- 17/2 , +/- 17

+/- 1/12 , +/- 1/6 , +/- 1/4 , +/- 1/3 , +/- 1/2 , +/- 1

That gives us 48 possible choices to start with and none of them work.

- Log in to reply to the answers

- sepiaLv 78 months ago
12x^50 + 89x^37 + 237x^21 + 122x^18 + 11x^14 - 342x^11 + 55x^9 - (12x)^7 +14x^5 + 37x^2 + 527 = 0

12x^50 + 89x^37 + 237x^21 + 122x^18 + 11x^14 + 55x^9 +14x^5 + 37x^2 + 527 = 342x^11 + (12x)^7

Exponent can't exceed 20

12(x^20 + x^20 + x^10) + 89(x^20 + x^17) + 237(x^5 + x^6)1 + 122x^18 + 11x^14 + 55x^9 +14x^5 + 37x^2 + 527 = 18x^7(19x^4 + 1990656)

- Log in to reply to the answers