F(x) = x^4 + 2x^2 - 24?

Question

okay, i get how to find a zero... one of them is -2....... but then how do i plug it into the equation???????????

Answer 1

f(x) = x^4 + 2x^2 - 24
f(x) = (x^2 + 6)(x^2 - 4)
f(x) = (x^2 + 6)(x - 2)(x + 2)

ANS : x = isqrt(6), -isqrt(6), 2, or -2

Answer 2

Another would be 2, since x is raised to an even numbered power, ^4 and ^2. Then it's 2^4 or 16, plus 2 times 2^2 or 8, minus 24. 16+8-24=0. The same works for -2.

The formula can be broken down to (x^2 +6)(X^2-4). Since there is no real number solution to (x^2 + 6)=0, you are left with (x^2 - 4)=0, with 2 and -2 as the solution.

Answer 3

Result
Solution 1 (real) -2

Solution 2 (real) +2

Solution 3 (complex) i*6^(1/2)

Solution 4 (complex) - i*6^(1/2)

How to plug in is very easy:
For example your one of the root is 2--> 2^4 + 2*2^2 -24 = 16 + 8 -24 =0
Similarly u can chk others

Answer 4

Divide the polynomial by x+2 to get a cubic. 2 is another root, so divide the cubic by x-2 to get a quadratic. Finish solving for the 2 remaining complex roots, i Sqrt(6) and - i Sqrt(6).