Use synthetic division to determine whether the number k is an upper or lower bound (as specified) for the real zeros of the function f. k = 4; f(x) = 2x^3 - 2x^2 - 3x - 5; Lower bound?

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Use synthetic division to determine whether the number k is an upper or lower bound (as specified) for the real zeros of the function f. k = 4; f(x) = 2x^3 - 2x^2 - 3x - 5; Lower bound?

Mathematics
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Yes or no are the answer choices
So, you have to solve for the zeros of the function. What are they?

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I'm honestly not sure how to find them
use demos graph to see where they cut x-axis
(2.2,0)
So would the answer be no since it's (2.2,0) and 4 would be a higher bound and not lower bound?
we don't have 0. please check
what do you mean? it cut the x-axis at 2.2
but you can use it. now find f(0) =?
replace 0 into f(x) , what do you get?
|dw:1442536923437:dw|
-5
ok, so f(0) =-5 < 0 , right? now f (2) =?
-3
So what does that mean
that mean from 0 to 2 you still not pass the x-axis, since f(x) <0 for both. Now test f(3)
22
yes, so from 2 to 3 , f(x) go from negative to positive, that is you have one zero on that interval, right?
Now, use synthetic,
forget it, short cut
okay haha
|dw:1442537533408:dw|
now, fill up
Again, the lesson didnt really explain this stuff so I'm not sure how to use that
you don't know synthetic?
not completely, looking it up now trying to learn real quick
|dw:1442537680182:dw|
|dw:1442537714109:dw|
now 4 *6 =?
24
|dw:1442537779766:dw| got it?
the last one, you do, the same
would it be 4*21 = 84 or 4*24 =96?
|dw:1442537857152:dw|
you take 4 * the bottom row, put the result on the middle row of the new column, then add them up to get the new number.
oh okay i see
that's easy then haha
|dw:1442538058710:dw|
ok, now, 4 >0 , right?
yes
look at the bottom row, they are 2,6,21,79 right?
yes
and all them are positive, right?
yes
Hence 4 is upper bound for the zeros
woah i see, a lot goes into it
so back to the original question the answer would be no
yup
ok thanks for all of the help, what would happen if one of them was negative?
you test, if k <0, the bottom line is +, -,+,- alternatively, then k is lower bound
learn from it http://www.howe-two.com/mathematicat/bound.html

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