List the possible zeros for the polynomial.h(x)=2x^4-3x^2-9​

Answer:[tex]x = \sqrt{3} [/tex][tex]x = - \frac{ \sqrt{6} }{2} [/tex]Step-by-step explanation:To solve, split the middle term like you would a quadratic. ac (2*-9) should be able to have multiples that add into b (-3).Which, it does: -6 and 3.2x^4-6x^2+3x^2-9. Now, you can factor. 2x^2(x^2-3)+3(x^2-3)You now have 2 factors to put together. The two in parentheses combine into one and the two outer terms become their own. (2x^2+3)(x^2-3)Now, set both equations equal to 0. 2x^2+3=0, subtract 32x^2=-3, divide by 2(I don't know where I went wrong here, but I still got the right answer?) x^2=-3/2 (you should get an imaginary number, but I know that's not the case. Assume the number is positive?) [tex]x = - \sqrt{6} \div 2[/tex]Do the same with the next, and you should get the square root of 3.