Find all real zeros and complexof each equation f(x)=(x^2+4)(x^2-9)

x = ± 3, x = ± 2i

Step-by-step explanation:

Given

f(x) = (x² + 4)(x² - 9)

To find the zeros let f(x) = 0, that is

(x² + 4)(x² - 9) = 0

Equate each factor to zero and solve for x

x² - 9 = 0 ( add 9 to both sides )

x² = 9 ( take the square root of both sides )

x = ± = ± 3

x² + 4 = 0 ( subtract 4 from both sides )

x² = - 4 ( take the square root of both sides )

x = ± = ± = ± × = ± 2i

Thus zeros are

x = - 3, x = 3 ← real

x = - 2i, x = 2i ← complex

i hope this is what you were looking for.

see attached graph

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6   = 0.00000180

1.80 x 10-4   = 0.000180 (bigger, or "less small", than the former one)

so

1.8 x 10-2       = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.