Question

Find all real zeros

f(x)=3x(x-9)^2(x-6)^2

Answer 1

they want you to find what x values make
3x(x-9)^2(x-6)^2=0

Answer 2

and the key idea is if *any term* is 0, the product is 0

Answer 3

ignoring the 3 which obviously can't be 0, you have 3 terms that might be 0, depending on what x is.

can you solve for x for each term, when the term is set equal to zero ?

Answer 4

Where do we start is my question

Answer 5

does this part make sense
if *any term* is 0, the product is 0
?

Answer 6

that is a yes or no question

Answer 7

yes

Answer 8

in
3x(x-9)^2(x-6)^2=0
list the 3 terms that could be zero:
x
(x-9)^2
(x-6)^2

Answer 9

ok

Answer 10

for each of those 3, we "solve" for x so that the term is 0
the first is easy , right ?
the other two, by taking the square root of both sides in the equation
\[ (x-9)^2 = 0 \]

Answer 11

9 for that one

Answer 12

yes, and because we can write (x-9)^2 as
(x-9)(x-9)=0
we have two terms (both become 0 when x is 9)
so point is, we call this a "repeated root"

Answer 13

or sometimes we say, 9 with multiplicity 2

I assume you can find x for the other two terms?

Answer 14

0,9,6 ?

Answer 15

yes, and if we want to be picky, 0, 9 repeated twice, 6 repeated twice

Answer 16

ok thank you!