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

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