Question

How do I find the roots and multiplicity of this:
y=4x^3-2x^2-2x?
I know that the answer is 0 and -2 with a multiplicity of 3 but i need to know how to do it

Answer 1

start by factoring the 'x' from all terms...

you'll get an easy quadratic as another factor :)

Answer 2

so that would be x(4x^2-2x-2)?

Answer 3

yep, notice you could factor out the 2 as well.

\(2x(2x^2 -x-1)\)

can you factor \(2x^2 -2x-1\) ?

Answer 4

can you factor the 2 out?

Answer 5

sure, we can.

Answer 6

\(\color{blue}{\text{Originally Posted by}}\) @hartnn
yep, notice you could factor out the 2 as well.

\(2x(2x^2 -x-1)\)

can you factor \(2x^2 -2x-1\) ?
\(\color{blue}{\text{End of Quote}}\)

i meant
can you factor \(2x^2 -x-1\) ?

Answer 7

I don't know how we could factor that any more because nothing can go into the 1

Answer 8

can you find 2 numbers whose sum is -1
and product is -2 ?

hint : only one of them is negative.

Answer 9

-2 and 1?

Answer 10

or -1 and 2?

Answer 11

-2 and 1 is correct!

so lets do factoring,
\(2x^2 -2x -x-1 \\ 2x(x+1) -1(x+1)\)

got that?
can you proceed?

Answer 12

so is that how we get the -2 as a root? How do we get the multiplicity?

Answer 13

-2 is not the root!

2x(x+1) -1 (x+1) = 0

gives,
(2x-1)(x+1) = 0

which means
2x-1 = 0 or x+1 = 0

so now can you find 2 other values of x ?
(we already had x=0 before)

Answer 14

Oh okay so we plug 0 into all of the x places?

Answer 15

we get x from

2x-1 = 0
so, 2x = 1
x =1/2

x+1 = 0
so, x = -1


so we have 3 roots of x as
-1, 0 , 1/2

got this?

Answer 16

yes i've got it

Answer 17

now about the mulitplicity,

it is how many times that particular root repeats.

for example, in 4,4,7.
The multiplicity of 4 = 2, because it repeats 2 times.
whereas the multiplicity of 7 is just 1.

Answer 18

Ohhhhh!! okay so because no number repeats it has a multiplicity of 1?

Answer 19

yup!
each root has the multiplicity of 1.

Answer 20

Yay!!!! Thank you soooo much for your help!!!! :D <3

Answer 21

welcome ^_^