Find the zeros of the polynomial function and state the multiplicity of each.
f(x) = 4(x + 7)2(x - 7)3

- ashking1

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- ashking1

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

- amoodarya

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

- phi

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

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## More answers

- ashking1

yes

- phi

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

- ashking1

0

- ashking1

3 times

- phi

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

- ashking1

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

- phi

it is easy to get confused.
4(x + 7)^2(x - 7)^3
we did (x-7) . what x makes that zero?

- ashking1

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 ?

- phi

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

- phi

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

- ashking1

no

- ashking1

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

- phi

yes

- ashking1

thanks btw can you help me with one more

- phi

ok

- ashking1

i need help understanding either the triangle or the theorem to answer this question
Expand the following using either the Binomial Theorem or Pascalâ€™s Triangle. You must show your work for credit.
(x - 5)5

- ashking1

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

- phi

please use the ^ otherwise it looks like (x-5) times 5
you mean \( (x-5)^5 \)
first, write it like this: \( (x + (-5) )^5 \)

- ashking1

sorry
ok

- phi

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 ?

- ashking1

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

- phi

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 ?

- ashking1

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

- phi

yes
keep going

- ashking1

a^2b^2

- phi

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

- ashking1

ya sorry thats what i ment

- phi

keep going, we need all the terms

- ashking1

then a^1B^4

- ashking1

a^0b^5

- phi

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.

- phi

start by writing this pattern
|dw:1437687651177:dw|

- phi

next we add a new row by adding 1 on each side, and adding the two numbers in the row above it. Like this
|dw:1437687710434:dw|

- ashking1

ok

- phi

now we keep adding rows until we get to a row with 6 entries... that will be the row we will use.
|dw:1437687767682:dw|

- phi

can you find the next row ?

- ashking1

|dw:1437687781164:dw|

- ashking1

i seen it before but how does it help for the problem

- phi

and we add those to our pattern
\[ (a+b)^5 = 1a^5 b^0 +5 a^4 b^1 + 10a^3b^2+10a^2b^3 + 5a^1 b^4 +1a^0 b^5 \]

- phi

we can simplify that a little... a^0 and b^0 are both 1. And we can leave off a coefficient of 1. so the pattern can be
\[ (a+b)^5 = a^5 +5 a^4 b + 10a^3b^2+10a^2b^3 + 5a b^4 +b^5 \]

- phi

now we can tackle your problem
\[ (x + (-5) )^5 \]
every where you see a in the pattern , replace it with x
and replace b with (-5)
can you do that ?

- ashking1

i changed everyhthing

- ashking1

everything

- phi

they probably want you to multiply out the (-5) to each exponent

- ashking1

ok

- phi

the first two terms should be
\[ x^5 - 25x^4 ...\]

- ashking1

sorry im writing everything on a piece of paper

- ashking1

+250x^3 is the next one right

- phi

yes

- ashking1

-1250x^2 is next one

- phi

yes

- ashking1

+6250x2-3,125

- phi

last -3125 is ok
but redo 5*x*(-5)^4

- ashking1

3125x

- phi

yes. so put it together for the final answer.

- ashking1

\[x^5-25x^4+250x^3-1250x^2+3125x-3125\]

- ashking1

thats the answer right

- phi

looks good

- ashking1

thank you so much

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