1)y=4x^3/2
2)f(x)=(3x^2+1)(2x^2+3)

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вопрос

To find the zeros of the first function, y=4x^(3/2), we set y equal to zero and solve for x:

0 = 4x^(3/2)
0 = x^(3/2)
0 = x^3

Therefore, x = 0 is the zero of the function.

To find the zeros of the second function, f(x) = (3x^2+1)(2x^2+3), we set f(x) equal to zero and solve for x:

0 = (3x^2+1)(2x^2+3)
0 = 3x^2 + 1 or 0 = 2x^2 + 3

For the first equation:
3x^2 + 1 = 0
3x^2 = -1
x^2 = -1/3
x = ± √(-1/3)

For the second equation:
2x^2 + 3 = 0
2x^2 = -3
x^2 = -3/2
x = ± √(-3/2)

Therefore, the zeros of the second function are x = ±i√(1/3) and x = ±i√(3/2).