Question

How do I give the multiplicity for each zero? Here is the Problem 2(x-5)(x+4)^2. Explain how you got the answer so I understand.

Answer 1

I would be able to do a problem like this f(x)=x(x+2)(x+4)

Answer 2

Refer to the attached plot.

Answer 3

I don't understand. What does that mean

Answer 4

\[2 (x-5) (x+4) (x+4)=2 x^3+6 x^2-48 x-160 \]Two of the roots are the same, -4 .
Other than that I can not help you.

Answer 5

hmmmmm...

Answer 6

??? I'll bump the question because I don't get it.

Answer 7

do you see how there is an exponent 2 on the factor (x+4)?

Answer 8

yes

Answer 9

\[(x+4)^2=0 \text{ will give you x=-4 multplicity 2 }\]

Answer 10

\[(x-5)^1 =0 \text{ will give you x=5 multplicity 1 }\]

Answer 11

What exactly does that mean.

Answer 12

that we have the same zero twice

Answer 13

because (x+4)^2=0
(x+4)(x+4)=0
set both factors equal to zero
x+4=0 or x+4=0
x=-4 or x=-4
do you see how we have the same zero twice, that is, we have the x=-4 twice?

Answer 14

we say the x=-4 occurs twice or you can use the term multiplicity 2

Answer 15

And that two at the front?

Answer 16

2 doesn't equal 0

Answer 17

the only things that can equal zero are (x+4)^2 or (x-5) factor that you have

Answer 18

What I mean is what I do with the 2 in 2(x-5)(x+4)^2

Answer 19

and what i mean is 2 doesn't equal 0
2=0 gives us no zeros because 2 will never be 0
\[2(x-5)(x+4)^2=0 \\ 2(x-5)(x+4)(x+4)=0 \\2=0\\x-5=0\\x+4=0\\x+4=0\]
2=0 gives us no zeros because 2 never equals 0
x-5=0 when x=5
x+4=0 when x=-4
x+4=0 when x=-4

Answer 20

So I ignore the 2???

Answer 21

yes because the factor 2 cannot be zero