Question

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

Answer 1

`(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))`