Question

Find all the zeroes of the equation.

–3x^4+ 27x^2 + 1200 = 0
Explain Please (:

Answer 1

divide by \(-3\) first and get
\[x^4-9x^2-400=0\]

Answer 2

by some miracle this actually factors as
\[(x^2-25)(x^2+16)=0\]

Answer 3

To make this question easier, you will notice that you can factor out a -3 out of the equation.
\[-3(x^{4} - 9x^{2} - 400) = 0\] You can then divide both sides by -3, which leaves you with \[x^{4} - 9x^{2} - 400 = 0\] Now this simplifies the problem a lot. Now this problem looks similar to something like \[x^{2} - 9x - 400 = 0\] You can use the same method you would to solve that to solve the problem you have. You can now try to factor the equation. You want two numbers that multiply to get -400, but added together to get -9. After some thought, I got 16 and -25. So now your equation factors out to\[(x^{2} + 16)(x^{2} - 25) = 0\] Notice how it looks similar to if you were to factor the quadratic, just you have x^2 in place of just x. Now you solve each one individually.

Answer 4

should be good from there
if you do not see that it factors, you can try replacing \(x^2\) by \(u\) and factor
\[u^2-9u-400=0\] but factoring is the key

Answer 5

take out the commom factor of -3

\[-3(x^4 -9x^2 - 400) = 0\]

then let u = x^2

so that you have a quadratic

so you need to solve

\[-3(u^2 - 9u - 400)=0\]

which factorises to

\[-3(u - 25)(u + 16) = 0\]

so u = 25 or -16

or x^2 = 25 or -16

now solve for x