Question

Find all the zeroes of 3x^4–11x^3–x^2+33x–24 if you known that 2 of is zeroes are √3 and -√3

Answer 1

Answer:

other 2 zeros are 1 and 8/3

Explanation:

$x = \sqrt{3} \: \: \: \: \: x = - \sqrt{3} \\ x - \sqrt{3} = 0 \: \: \: \: x + \sqrt{3} = 0 \\ (x - \sqrt{3} )(x + \sqrt{3} ) = 0 \\ {x}^{2} - 3 = 0 \\ dividing \: 3 {x}^{4} - 11 {x}^{3} - {x}^{2} + 33x - 24 \: by \: {x}^{2} - 3 \\ \frac{3 {x}^{4} - 11 {x}^{3} - {x}^{2} + 33x - 24 }{ {x}^{2} - 3} = 3 {x}^{2} - 11x + 8 \\ \\ 3 {x}^{2} - 11x + 8 = 0 \\ 3 {x}^{2} - 3x - 8x + 8 = 0 \\ 3x(x - 1) - 8(x - 1) = 0 \\ (x - 1)(3x - 8) = 0 \\ x - 1 = 0 \: \: \: 3x - 8 = 0 \\ x = 1 \: \: \: \: \: x = \frac{8}{3}$

Answer 2

Answer:

just copy it man ok but try it again urself