ashking1
  • ashking1
Find the zeros of the polynomial function and state the multiplicity of each. f(x) = 4(x + 7)2(x - 7)3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ashking1
  • 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
  • amoodarya
do you mean ?\[f(x) = 4(x + 7)^2(x - 7)^3\]
phi
  • 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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ashking1
  • ashking1
yes
phi
  • phi
what happens if you erase the x and put in 7 in its place in (x-7) what number do you get ?
ashking1
  • ashking1
0
ashking1
  • ashking1
3 times
phi
  • phi
yes. they use the word "multiplicity" so look for an answer that says 7 multiplicity 3
ashking1
  • ashking1
-7, multiplicity 3; 7, multiplicity 2 so would this be the answer
phi
  • phi
it is easy to get confused. 4(x + 7)^2(x - 7)^3 we did (x-7) . what x makes that zero?
ashking1
  • 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
  • phi
(x-7) is zero when x=7 and multiplicity 3 because we have (x-7)^3
phi
  • phi
you picked *** 7, multiplicity 2 so would this be the answer***
ashking1
  • ashking1
no
ashking1
  • ashking1
it would be -7, multiplicity 2; 7, multiplicity 3
phi
  • phi
yes
ashking1
  • ashking1
thanks btw can you help me with one more
phi
  • phi
ok
ashking1
  • ashking1
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
ashking1
  • ashking1
like you dont have to tell me the answer more like help me understand how to get there
phi
  • 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
  • ashking1
sorry ok
phi
  • 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
  • ashking1
so it will be (5+0)^5= (4)(1).....
phi
  • 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
  • ashking1
would the next one be like a^3b^2
phi
  • phi
yes keep going
ashking1
  • ashking1
a^2b^2
phi
  • 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
  • ashking1
ya sorry thats what i ment
phi
  • phi
keep going, we need all the terms
ashking1
  • ashking1
then a^1B^4
ashking1
  • ashking1
a^0b^5
phi
  • 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
  • phi
start by writing this pattern |dw:1437687651177:dw|
phi
  • 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
  • ashking1
ok
phi
  • 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
  • phi
can you find the next row ?
ashking1
  • ashking1
|dw:1437687781164:dw|
ashking1
  • ashking1
i seen it before but how does it help for the problem
phi
  • 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
  • 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
  • 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
  • ashking1
i changed everyhthing
ashking1
  • ashking1
everything
phi
  • phi
they probably want you to multiply out the (-5) to each exponent
ashking1
  • ashking1
ok
phi
  • phi
the first two terms should be \[ x^5 - 25x^4 ...\]
ashking1
  • ashking1
sorry im writing everything on a piece of paper
ashking1
  • ashking1
+250x^3 is the next one right
phi
  • phi
yes
ashking1
  • ashking1
-1250x^2 is next one
phi
  • phi
yes
ashking1
  • ashking1
+6250x2-3,125
phi
  • phi
last -3125 is ok but redo 5*x*(-5)^4
ashking1
  • ashking1
3125x
phi
  • phi
yes. so put it together for the final answer.
ashking1
  • ashking1
\[x^5-25x^4+250x^3-1250x^2+3125x-3125\]
ashking1
  • ashking1
thats the answer right
phi
  • phi
looks good
ashking1
  • ashking1
thank you so much

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