Question

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

Answer 1

`x = √(3)`

`x = - ( √(6) )/(2)`

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 3

2x^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?)

`x = - √(6) / 2`

Do the same with the next, and you should get the square root of 3.