Question

if the zeros of your x⁴+x³-5x²-3x+6 are root 3 and - root 3 then find the other two zeros​

Answer 1

$- 2,1$

Step-by-step explanation:

$Given, \\ {x}^{4} + {x}^{3} - 5 {x}^{2} - 3x + 6 \\ x = \sqrt{3} \: \: and \: \: - \sqrt{3} \\ then \: \: \: (x - \sqrt{3} )(x + \sqrt{3} ) = {x}^{2} - 3 \\ now \: \\ \frac{{x}^{4} + {x}^{3} - 5 {x}^{2} - 3x + 6}{ {x}^{2} - 3} = {x}^{2} + x - 2 \\ {x}^{2} + x - 2 = 0 \\ {x}^{2} + 2x - x - 2 = 0 \\ x(x + 2) - 1(x + 2) = 0 \\ (x + 2)(x - 1) = 0 \\ x = - 2 \: \: and \: \: x = 1 \\ all \: \: roots \: \: are \: \: \sqrt{3}, \: - \sqrt{3} , \: - 2 \: \: and \: \: 1$