Given that 7 is a zero of h(x)=x^3-7x^2-3x+21, find the other zeros.

x= 7    x = ±√3

Step-by-step explanation:

h(x)=x^3-7x^2-3x+21

Factor by grouping

h(x) = x^2( x-7)  -3 (x-7)

Factor out an (x-7)

h(x) = (x-7)  ( x^2 -3)

Using the zero product property

0= (x-7)  ( x^2 -3)

x-7 =0     x^2 -3 =0

x=7     x^2 =3

x= 7    x = ±√3

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~