Question

f(x) = x2 - 16 and g(x) = x+4. Find F over G of and its domain

Answer 1

\[\frac{ f(x) }{ g(x) }\]
is this what you mean?

Answer 2

yeah

Answer 3

\[\frac{ (x^2 -16) }{ (x+4) }\]
you can factor the numerator in a way so as to get two expressions one of which will be (x+4) which will cancel out with the denominator

Answer 4

so in other words (x+4)(?) = (x^2 - 16)

Answer 5

4x and 16x^2 right

Answer 6

im sorry i don't quite understand your expression can you write it out?

Answer 7

im not getting it

Answer 8

okay do you know how to factor \[(x^2-16)\]

Answer 9

x^2 - 16 is the difference of 2 squares
x^2 is a perfect square and so is 4

Answer 10

have you factored something like that before?

Answer 11

no

Answer 12

are you familiar with the difference of squares method (its the way to factor expresions like these )

Answer 13

yeah but its a real number right

Answer 14

basically it means that if your constant (in this case the 16) is a perfect square you can factor the expression by multiplying two expressions which's constant is the perfect square of your original constant

Answer 15

okay im over complicating this XD

Answer 16

okay do you know what \[\sqrt{16} = ?\]

Answer 17

2*8

Answer 18

umm no , what number squared equals 16

Answer 19

256

Answer 20

no thats 16 squared

Answer 21

no what number when multiplied by itself equals 16

like 3 * 3 = 9 So 3 is the square root of 9.

Answer 22

4

Answer 23

right

Answer 24

correct which means that \[\sqrt{16} = 4\]
and because 4 is a whole number we call 16 a perfect square

Answer 25

okay thanks