Question

Find other zeros of 2x^4 -9x^3+5x^2 +3x if two of its zero are 2+root 3 and2-root 3

Answer 1

$x = 2 + \sqrt{3} \\( x - 2 - \sqrt{3}) \: is \: a \: factor \\ x = 2 - \sqrt{3} \\( x - 2 + \sqrt{3} ) \: is \: another \: factor \\ multiply \: both \: factors \\ = {x}^{2} - 4x + 4 - 3 \\ = {x}^{2} - 4x + 1 \\ 2 {x}^{4} - 9 {x}^{3} + 5 {x}^{2} + 3x = 0 \\ x(2 {x}^{3} - 9 {x}^{2} + 5x + 3) = 0 \\ x = 0 \: is \: 3rd \: zero \\ {x}^{2} - 4x + 1)2 {x}^{3} - 9 {x}^{2} + 5x + 3(2x - 1\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2 {x}^{3} - 8 {x}^{2} + 2x \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: + \: \: \: \: \: \: \: - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - {x}^{2} + 3x + 3 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - {x}^{2} + 4x + 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: + \: \: \: \: \: - \: \: \: \: \: \: \: \: - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - x + 2$
is remainder.
There is something wrong in this question,might you enter wrong polynomial,because if those are zeros of polynomial,then there must be 0 in the remainder.
check your question again.
the method I had mentioned you