Question

Find the zeros of the polynomial function and state the multiplicity of each.

f(x) = 4(x + 7)2(x - 7)3

Answer 1

these are the choices

4, multiplicity 1; -7, multiplicity 3; 7, multiplicity 3

-7, multiplicity 3; 7, multiplicity 2

4, multiplicity 1; 7, multiplicity 1; -7, multiplicity 1

-7, multiplicity 2; 7, multiplicity 3

can someone help me please

Answer 2

do you mean ?\[f(x) = 4(x + 7)^2(x - 7)^3\]

Answer 3

you should put in ^ to show exponents. I assume this is
f(x) = 4(x + 7)^2(x - 7)^3
\[ f(x) = 4(x + 7)^2(x - 7)^3 \]

Answer 4

yes

Answer 5

what happens if you erase the x and put in 7 in its place in (x-7)
what number do you get ?

Answer 6

0

Answer 7

3 times

Answer 8

yes. they use the word "multiplicity"
so look for an answer that says 7 multiplicity 3

Answer 9

-7, multiplicity 3; 7, multiplicity 2 so would this be the answer

Answer 10

it is easy to get confused.
4(x + 7)^2(x - 7)^3

we did (x-7) . what x makes that zero?

Answer 11

what do you mean by that

like its the second x the one that says in (x-7) and if x is 7 it will equal 0 right ?

Answer 12

(x-7) is zero when x=7 and multiplicity 3 because we have (x-7)^3

Answer 13

you picked *** 7, multiplicity 2 so would this be the answer***

Answer 14

no

Answer 15

it would be
-7, multiplicity 2; 7, multiplicity 3

Answer 16

yes

Answer 17

thanks btw can you help me with one more

Answer 18

ok

Answer 19

i need help understanding either the triangle or the theorem to answer this question

Expand the following using either the Binomial Theorem or Pascal’s Triangle. You must show your work for credit.

(x - 5)5

Answer 20

like you dont have to tell me the answer more like help me understand how to get there

Answer 21

please use the ^ otherwise it looks like (x-5) times 5
you mean \( (x-5)^5 \)

first, write it like this: \( (x + (-5) )^5 \)

Answer 22

sorry

ok

Answer 23

next, we start with this pattern
\[ (a+b)^5 = a b + a b + a b + ...\]
make the exponent of the a in the first term 5, and the exponent on b 0
(the exponents will *always* add up to 5)
the next a will have an exponent of 4 (and the b will have 1)
can you write out the a b pattern for (a+b)^5 ?

Answer 24

so it will be (5+0)^5= (4)(1).....

Answer 25

no. I am trying to say that
\[ (a+b)^5 = a^5 b^0 + a^4 b^1 + ... \]
(there are also numbers we have to put in, but first we need to learn how to write down this pattern)

can you finish writing out the full pattern ?

Answer 26

would the next one be like a^3b^2

Answer 27

yes
keep going

Answer 28

a^2b^2

Answer 29

you mean a^2 b^3 (remember the exponents add up to 5)
or the a goes down, the one on b goes up

Answer 30

ya sorry thats what i ment

Answer 31

keep going, we need all the terms

Answer 32

then a^1B^4

Answer 33

a^0b^5

Answer 34

yes. so we have (so far)
\[ (a+b)^5 = a^5 b^0 + a^4 b^1 + a^3b^2+a^2b^3 + a^1 b^4 +a^0 b^5\]
now we have to put in coefficients (numbers) in front of each term.
we use Pascal's triangle to find these numbers.

Answer 35

start by writing this pattern

Answer 36

next we add a new row by adding 1 on each side, and adding the two numbers in the row above it. Like this