How do you find all the zeros of #x^5-2x^4-4x^3-10x^2-21x-12#?

1 Answer
May 24, 2016

Answer:

#(x+1)^2(x-4)(x+i sqrt(3))(x+i sqrt(3))#

Explanation:

The polynomial has constant coefficient #-12# which is the product of its roots.

Trying the sequence of #-12# factors given by

#{pm 1,pm2, pm 3, pm 4, pm 6, pm 12}#

we get

#p(-1) = p(4) = 0#

then at last we have found two roots ( keep in mind that some of this roots can be multiple). So we can write

#p(x) = x^5 - 2 x^4 - 4 x^3 - 10 x^2 - 21 x - 12#

and also

#p(x) - (x+1)(x-4)(x^3+a x^2+b x+c) = 0# for all #x#

Taking the coefficients equal to zero we have

#((12 - 4 c=0), (21 - 4 b - 3 c=0),(10 - 4 a - 3 b + c=0), (-3 a + b=0), (-1 + a=0))#.

Solving for #a,b,c# we get

#(a = 1, b = 3, c =3)#

Focusing now in the roots of

#x^3+x^2+3x+3=0#

and having in mind the ideas about the composition of the constant term we find that #x = -1# is a root so we can write

#x^3+x^2+3x+3 = (x+1)(x^2+d x + e)#

proceeding as before we find #d = 0, e = 3#. Putting all together we have

#p(x)=(x+1)^2(x-4)(x+i sqrt(3))(x+i sqrt(3))#