Question

p(x)=x^3(x+2)(x-5)^2

Answer 1

find the multiplicity and zeros of each

Answer 2

it is the exponent

Answer 3

\[x^3(x+2)(x-5)^2=0\] one factor is \(x^3\) which is zero if \(x\) is
the "multiplicity " of that zero is three, because of the cube

Answer 4

what do you do with the other exponent?

Answer 5

so we would say "0 is a zero of multiplicity 3"
now \(x+2=0\) if \(x=-2\) so \(-2\) is also a zero

Answer 6

the multiplicity of that zero is one, because the factor is just \((x+2)\)
you would say "\(-2\) is a zero of multiplicity one"

Answer 7

how about \((x-5)^2\) ?

Answer 8

yeah whats that all about

Answer 9

lol
solve \(x-5=0\) what do you get?

Answer 10

is there any synthetic division/long division required in this problem?

Answer 11

so the other zero is 5?

Answer 12

no you do this problem only with your eyeballs
no computation necessary at al
yes, the other zero is 5

Answer 13

what is the "multiplicity" of that zero?

Answer 14

5? :/

Answer 15

wait! 1!

Answer 16

no

Answer 17

the zero is 5
for the multiplicity recall that the factor is \((x-5)^2\)

Answer 18

\[\large (x-\color{red}{5})^{\color{blue}2^{\text{ multiplcity is here}}}\]

Answer 19

oh so the multiplicity is 2?

Answer 20

right

Answer 21

see, you just look at it, it is displayed for you
nothing at all to compute

Answer 22

this is so sketchy. my teacher threw this in our homework as a curve ball because in the examples in class we had to divide and then factor out all this stuff and do the p/q thing to find the zeros

Answer 23

yeah, but this problem starts at the end
it is already in factored form, so there is nothing to do

Answer 24

i.e. the factoring has already been done for you

Answer 25

awesome that makes sense!

Answer 26

good luck

Answer 27

now i have to draw a graph

Answer 28

ok it goes like this
you have zeros at \(-2,0,5\) that is where the graph hits the \(x\) axis

Answer 29

this is where the "multiplicity" comes in
because the zero at \(-2\) has multiplicity 1, the curve crosses the \(x\) axis at \(-2\)

Answer 30

that is because \(1\) is an odd number
since \(3\) is odd as well, the curve crosses the \(x\) axis at \(0\)

Answer 31

but \(2\) is even, so the curve does not cross the \(x\) axis at \(5\) it only touches it there