Question
-3x^4+27x^2+1200=0
Answers
Steve
-3(x^2+16)(x^2-25) = 0
That better?
That better?
the world forgot
thats the step i got to before i dont know what to do
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.
-3(x^2+16)(x-5)(x+5) = 0
the product if zero if any of the factors is zero.
the world forgot
im asking what the zeros for the equation are
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.
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.
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
0=3x^4-27x^2-1200
(3x²+48)(x²-25)=0
3(x²+16)(x+5)(x-5)=0