Question

y = (x – 2)^2(x – 1)
y = 3x^3 – 3x
Find the zeroes?

Answer 1

to find the zeros, put y= 0 in each of them

Answer 2

So
y = (x – 2)^2(x – 1)
(x-2)^2(x-1)=0
I don't know what to do next for this one.
y = 3x^3 – 3x
3x^3-3x=0
This one divide by 3
x^3-x=0
Then I'm stuck again

Answer 3

ok, when we have \((x-a)(x-b) =0 \)
we can say that either
\(x-a = 0 \quad or \quad x-b =0 \)

Answer 4

Yeah I got that.

Answer 5

so if we have
\((x-2)^2(x-1) = 0 \)
what can we say ?

Answer 6

(x-2)^2=0; x=? The square throws me off.
(x-1)=0; x=1

Answer 7

ok,
(x-a)^2 = 0
means
x-a = 0 or x-a = 0
so, x = a or a
which means x= a with the multiplicity of 2 !

Answer 8

is that new to you ?

Answer 9

I'm still a little confused
So I have
(x-2)^2=0
x-2=0
or
x-2=0?
They'd be the same x=2 according to what you wrote that's how I should do it.

Answer 10

yes, same roots or EQUAL roots
instead of saying, x = 2,2
we say that x= 2 with the multiplicity of 2!

Answer 11

Oh ok that kind of makes sense.
So my answer would be
x=1, 2 multiplicity of 2

Answer 12

yes, thats correct :)

Answer 13

Okay great!

Answer 14

for x^3-x = 0
first factor out x from the left

Answer 15

So
x(x^2-1)=0?

Answer 16

correct!

so,
x= 0 or x^2 -1 =0

can you solve x^2 -1 =0 ??

Answer 17

its easy :)

Answer 18

(x-1)(x+1)=0
x-1=0
x+1=0
x=0,1,-1

Answer 19

correct!

Answer 20

good work :)

Answer 21

Thank you!

Answer 22

most welcome ^_^