Question

Identify the roots and state the multiplicity

x^6 - 12x^4 + 48x^2 -64 = 0

Answer 1

i put it in the calculator and I see that the zeros are at 2 and -2
i am not clear on the multiplicity of this
@Chlorophyll

Answer 2

hi @vishal kothari can you help me understand this?

Answer 3

@saifoo.khan

Answer 4

Sorry i dont really know how to do it.

But what i think is, we have to suppose the values for let's say x^2
x^2 =y

Then equation will look something like,

y^3 - 12 y^2 + 48y - 64 = 0

Answer 5

i think i solved it. lol
Now factor this out,

y^3 - 12 y^2 + 48y - 64 = 0
(y-4)(y-4)(y-4) = 0 <--- used a calc here.

you have y = 4.

Now use this value here; x^2 =y

Answer 6

@JuanitaM , does it make any sense?

Answer 7

well oddly enough the zeroes are 2 and -2 not 4. thanks anyway

Answer 8

Yes! right.

Answer 9

What saifoo is saying is that \(y=x^2=4\) thus, \(x=\pm 2\) . Then, since you have degree 6, and each root is either \(\pm 2\), both roots have multiplicity 3.

Answer 10

i dont get the degree thing.. but i was correct. :D

Answer 11

yes u were :)

Answer 12

To find the multiplicity more explicitly, take the function\[(x^2-4)(x^2-4)(x^2-4)=0\]Then for each root \(x^2=4\), of which there are three, you have two roots; \(+2, -2\). Since each of these appears three times, each has multiplicity 3.

Answer 13

Ah, prefect.

Answer 14

great - but how did you off the back see x^2 -4

we graphed and that is how i knew the zeros was +-2. I misread the answer so each term is amultiplicy of 3 with a total of 6

Answer 15

When saifoo wrote \((y-4)(y-4)(y-4)\), this is equal to \((x^2 -4)(x^2 -4)(x^2 -4)\) since \(y=x^2\).

That's where I got the \((x^2 -4)(x^2 -4)(x^2 -4)\) from.

Answer 16

i think it is thee cube of first and last terms -