Question

Find all the zeroes of the polynomial p(x) = x4 – 7x³ + 9x² + 13x – 4, if two of its zeroes are (2 +√ 3) and ( 2 - √ 3).

Answer 1

polynomial of the two zeroes
(x - (2+rt(3))(x - (2-rt(3)) = x^2 -4x +1

divide it
(x^4 - 7x^3 + 9x^2 + 13x – 4) / (x^2 -4x +1)
= x^2 - 3x -4
= (x-4)(x+1)
x=-1, 4 ... the remaining zeroes

Answer 2

All the zeroes of the polynomial p(x) are:

`-1\ ,\ 4\ ,\ (2+√(3))\ ,\ (2-√(3))`

Explanation:

We are given a polynomial expression p(x) as:

`p(x)=x^4-7x^3+9x^2+13x-4`

Also two of the zeros of the polynomial are:

`2+√(3)\ and\ 2-√(3)`

i.e. the polynomial could be factored as follows:

`x^4-7x^3+9x^2+13x-4=(x-(2+√(3)))(x-(2-√(3)))q(x)`

where q(x) is a polynomial of degree 2

i.e.

`x^4-7x^3+9x^2+13x-4=(x-(2+√(3)))(x-(2-√(3)))q(x)`

i.e.

`q(x)=(x^4-7x^3+9x^2+13x-4)/(x^2-4x+1)`

Hence, we get the value of q(x) as:

`q(x)=x^2-3x-4\\\\\\i.e.\\\\\\q(x)=x^2-4x+x-4\\\\\\i.e.\\\\\\q(x)=x(x-4)+1(x-4)\\\\\\i.e.\\\\\\q(x)=(x+1)(x-4)`

Hence, our polynomial could be represented in terms of linear factors as:

`x^4-7x^3+9x^2+13x-4=(x-(2+√(3)))(x-(2-√(3)))(x+1)(x-4)`

i.e. the other two zeros are: -1 and 4