Asked by kaley
find all the zeros of the polynomial function
F(x)= x^4+4x^3-6x^2-36x-27
F(x)= x^4+4x^3-6x^2-36x-27
Answers
Answered by
drwls
There are websites that solve such root-finding questions automatically, but they are not very instructive.
If there are integer roots, they must be even divisors of 27: +/- 1, 3, 9 or 27.
That is a consequence of the "rational roots theorem", which you should learn.
One such root is -1, so x+1 is a factor.
The other factor is
(x^4+4x^3-6x^2-36x-27)/(x+1)
= x^3 +3x^2 -9x -27
(obtained with polynomial long division)
x = 3 is clearly another root, so (x-3) is another factor. Divide the cubic by (x-3) and you get
x^2 + 6x +9 = 0
which can be factored to give
(x+3)^2 = 0
That means x = -3 is a double root.
The roots are -3, +3 and -1
If there are integer roots, they must be even divisors of 27: +/- 1, 3, 9 or 27.
That is a consequence of the "rational roots theorem", which you should learn.
One such root is -1, so x+1 is a factor.
The other factor is
(x^4+4x^3-6x^2-36x-27)/(x+1)
= x^3 +3x^2 -9x -27
(obtained with polynomial long division)
x = 3 is clearly another root, so (x-3) is another factor. Divide the cubic by (x-3) and you get
x^2 + 6x +9 = 0
which can be factored to give
(x+3)^2 = 0
That means x = -3 is a double root.
The roots are -3, +3 and -1