# need help finding the complex roots of a 6th degree polynomial

• April 6th 2010, 03:55 PM
heretostay
need help finding the complex roots of a 6th degree polynomial
hello, i have a homework problem that is giving me all kinds of trouble. If any one could help it would be greatly appreciated.

the equation is $12x^6-65x^5-69x^4+159x^3-21x^2+224x+60$

I hope I posted that in an acceptable format.
So far i have the real roots which are 6, -2, 1 2/3, -.25

I am not sure where to go from here to find the complex roots...

• April 6th 2010, 05:30 PM
harish21
Quote:

Originally Posted by heretostay
hello, i have a homework problem that is giving me all kinds of trouble. If any one could help it would be greatly appreciated.

the equation is $12x^6-65x^5-69x^4+159x^3-21x^2+224x+60$

I hope I posted that in an acceptable format.
So far i have the real roots which are 6, -2, 1 2/3, -.25

I am not sure where to go from here to find the complex roots...

$12x^6-65x^5-69x^4+159x^3-21x^2+224x+60 = 0$

$(x-6) (x+2) (3x-5) (4x+1) (x^2+1) = 0$

So the real solutions are:

$x = -2 , \frac{-1}{4}, \frac{5}{3}, 6$

and the complex solutions are

$x = i, -i$

because:
$(x^2+1) = 0 \rightarrow {x^2} = -1 \rightarrow x = \pm i$
• April 6th 2010, 05:38 PM
heretostay
thanks for your help, i am not exactly sure how your factored that problem, if its not too much trouble, would you be able to explain that to me?
• April 7th 2010, 03:51 AM
HallsofIvy
You told us that " the real roots are 6, -2, 1 2/3, -.25". That tells us that x minus each of those: x- 6, x-(-2)= x+ 2, x- 5/3, and x- -.25= x- 1/4 are factors. Since x- 5/3= (1/3)(3x- 5) is a factor so is 3x- 5. Since x- 1/4= (1/4)(4x- 1) is a factor so is 4x- 1.

That is: the polynomial is $(x- 6)(x+2)(3x-5)(4x-1)(ax^2+ bx+ c)$. You can find the coefficients of the quadratic factor, a, b, and c, by multiplying all of that out and comparing to the original polynomial or by multiplying out the linear factors and dividing into the original polynomial.