Answer:
All the zeroes of the polynomial p(x) are:
\(-1\ ,\ 4\ ,\ (2+\sqrt{3})\ ,\ (2-\sqrt{3})\)
Step-by-step 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:
\(2+\sqrt{3}\ and\ 2-\sqrt{3}\)
i.e. the polynomial could be factored as follows:
\(x^4-7x^3+9x^2+13x-4=(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))q(x)\)
where q(x) is a polynomial of degree 2
i.e.
\(x^4-7x^3+9x^2+13x-4=(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))q(x)\)
i.e.
\(q(x)=\dfrac{x^4-7x^3+9x^2+13x-4}{x^2-4x+1}\)
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:
\(x^4-7x^3+9x^2+13x-4=(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))(x+1)(x-4)\)
i.e. the other two zeros are: -1 and 4
