Asked by the world forgot

-3x^4+27x^2+1200=0



Answered by Steve
-3(x^2+16)(x^2-25) = 0
That better?
Answered by the world forgot
thats the step i got to before i dont know what to do
Answered by Steve
what? Have you forgotten your Algebra I?

-3(x^2+16)(x-5)(x+5) = 0

the product if zero if any of the factors is zero.
Answered by the world forgot
im asking what the zeros for the equation are
Answered by Steve
the zeros of the function are where any factor is zero. What are you doing in pre-cal???

when is -3 = 0? never
when is x^2+16 = 0? never (x^2 is always positive, so x^2+16 cannot be zero)

x-5=0 when x=5
x+5=0 when x = -5

So, the zeros are x = ±5

Looks like you have some serious review to do.
Answered by unowen
-3x^4+27x^2+1200=0
0=3x^4-27x^2-1200
(3x²+48)(x²-25)=0
3(x²+16)(x+5)(x-5)=0
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