finding all zeros of the polynomial

Question

find all zeros of p(x)=6x^7+35x^6-169x^5-612 x^4+1823x^3+1215x^2-2758 x-980 where p (1+radical 5)=0

Answer 1

You can try to take the derivative, then factor the derivative and set each part equal to 0 and solve for it, after that you would have to do a first derivative test to see which ones are maxes/mins or you can use a graphing calculator graph that beast and see where it touches the x-axis which is what I would do.

Too lazy to do that because its going to be messy for this function

Answer 2

Find all zeros of p(x)=6x^7+35x^6-169x^5-612x^4+1823x^3+1215x^2-2758x-980
where p (1+radical 5)=0.

Let 6x^7+35x^6-169x^5-612x^4+1823x^3+1215x^2-2758x-980 = 0

One root is x = 1 + rt(5).  

Since all the coefficients and the constant value, 980, are integers, this suggests that one other root is x = 1 – rt(5), these two roots being the roots of a quadratic expression.
Multiplying them together gives (x – (1+rt(5))(x – (1-rt(5)) = x^2 – 2x – 4.

Dividing p(x) by (x^2 – 2x – 4) gives us,
(x^2 – 2x – 4)(6x^5+47x^4-51x^3-526x^2+567x+245 ) = 0

Now let us look at the quintic polynomial.
6x^5+47x^4-51x^3-526x^2+567x+245 = 0   ------------- (1)

This is a 5th degree polynomial and as such has 5 roots and 5 factors. The polynomial can thus be written as,
(ax + b)(cx + d)(ex + f)(gx + h)(ix + j) = 0

Now the coefficient of x^5 in (1)  is 6.
And the product acegi = 6.
But 6 only has the factors 2 and 3.
So let a = 2, c = 3 and e = g = i = 1. Then
(2x + b)(3x + d)(x + f)(x + h)(x + j) = 0

Also, the product of all the constants in the factors = the constant value in the polynomial (1).
i.e. bdfhj = 245 = 5 * 7^2.
So, values for b, d, f, h, j are 1 or 5 or 7.
So try root values for x as +/- 1, +/- 5, +/- 7, +/- ½, +/- 5/2, +/- 7/2, +/- 1/3, +/- 5/3, +/- 7/3.

Substituting for the above possible root values into p(x) gives us that
x = - 7 is a root.
x = + 5/2 is a root.
x = -1/3 is a root.

So, factors are: (x + 7), (2x – 5), (3x + 1).

Dividing the quintic by the above factors gives us,
(x + 7)(2x – 5)(3x + 1)(x^2 + 3x – 7)(x^2 – 2x –  4) = 0

The final quadratic is x^2 + 3x – 7, which is solved using the quadratic formula, and gives the final two roots as,
x = -3/2 + (i/2)*rt(19), and x = -3/2 – (i/2)*rt(19)

So, the seven roots of the original polynomial p(x) are,

x = -7, x = 5/2, x = -1/3, x = -3/2 + (i/2)*rt(19), x = -3/2 – (i/2)*rt(19), x = 1 + rt(5), x = 1 – rt(5)