Question

obtain all zeroes of p(x)=x^4-3x^3-x^2+9x-6

Answer 1

to find the zeros of the given equation ,i use the synthetic division method.

Answer 2

Answer-

The zeros of the given polynomial are 1, 2, √3, -√3

Solution-

Given,

$p(x)=x^4-3x^3-x^2+9x-6$

For calculating zeros or roots,

$\Rightarrow p(x)=0$

$\Rightarrow x^4-3x^3-x^2+9x-6=0$

$\Rightarrow x^4-2x^3-x^3+2x^2-3x^2+6x+3x-6=0$

$\Rightarrow x^3(x-2)-x^2(x-2)-3x(x-2)+3(x-2)=0$

$\Rightarrow (x-2)(x^3-x^2-3x+3)=0$

$\Rightarrow (x-2)(x^2(x-1)-3(x-1))=0$

$\Rightarrow (x-2)(x^2-3)(x-1)=0$

$\Rightarrow x=1,2,\pm \sqrt{3}$