6x^4 + x^3 + 10x^2 + x + 6 = 0 solve for x.?

Question

Could you also please show the process? Thank you!

It's a homework assignment.

The question just says: "Solve the following equations". So I assume I'm supposed to be solving for x right? Why would the teacher give an impossible equation. :(

maybe its a trick question or mistake :P

Answer 1

divide by x^2
6(x^2+(1/x^2))+ (x)+(1/x)+10 = 0
put x+(1/x)= t

then solve for t and then solve for x
you'll find that there are no roots.!

Answer 2

First of all it is a wrong question. No products of four factors will give 6x^4 + x^3 + 10x^2 + x + 6 = 0 as it doesn't match with the y-intercept i.e 6.

It is possible to solve this type of equation. Many people think only quadratic or linear equations can be solved. Using a proper method it can solved.

Observe: c = 6. That means from general equation

(x+a)(x+b)(x+c)(x+D)
c=a*b*c*d
a*b*c*d=6

take factors of 6 which multiply four times to give c value as 6.

Factors are: -1,-2,-3,-6,1,2,3,6
Only one combination will work. (Select 4 numbers from the list which give product of 6 and divide to give reminder 0. (Long process)

For example: If we take value 1 i.e (x+1). Now divide the main equation by (x+1) to get a cubic equation. If the reminder is 0, then the factor is correct. Then one of the values will be x = -1

Answer 3

because of the fact that a number of those are superior mutually, making each area equivalent 0 makes the full equation equivalent 0 If x=6, then (x-6) equals 0, and multiplying 0 by skill of something equals 0. So x equals 6, 17 or -4 is actual. yet which one? in case you in basic terms artwork the undertaking out with any of those values for x, you will get 0. So all are maximum remarkable.

Answer 4

This problem is simple.
look at the coefficients

they are 6,1,10,1,6
it is symetric

so (x+1/x) comes as a part of a factor

devide by x^2

6x^2+x+10+ 1/x+6/x^2
= 6(x^2+1/x^2)+ (x+1/x) + 10

let x+1/x = t or tx = x^2+1
so x^2+2/x^2 = t^2 -2
so expression/x^2 = 6(t^2-2) + t + 10
= 6t^2 +t - 2
= 6t^2+ 4t-3t-2
= 2t(3t+2) -(3t+2)
= (2t-1)(3t+2)
expression = x^2(2t-1)(3t+2)
= (2tx -x)(3tx + 2x)
= (2x^2-2-x)(3x^2+3 + 2x)
= (2x^2-x-2)(3x^2+2x + 3)
both being binomial can be attempted as below

(2x^2-x-2) = (2x+1)(x-2)
(3x^2+2x+3 cannot be factored
so (2x+1)(x-2)(3x^2+2x+3)

Answer 5

impossible
the first 4 terms must equal -6
if x is positive this isnt possible
if x is negative the second and 4th term would be negative but the first and third term are too large respectively to the next term that would be negative. meaning the 6x^4+x^3+10x^2+x has to be positive

Answer 6

impossible, no matter what value there is for x, the number will be no less than six. (0,6)

Answer 7

the answer to your problem is: no solution