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?

- anonymous

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- anonymous

Yes or no are the answer choices

- anonymous

@Loser66

- Loser66

So, you have to solve for the zeros of the function. What are they?

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## More answers

- anonymous

I'm honestly not sure how to find them

- Loser66

use demos graph to see where they cut x-axis

- anonymous

(2.2,0)

- anonymous

So would the answer be no since it's (2.2,0) and 4 would be a higher bound and not lower bound?

- Loser66

we don't have 0. please check

- anonymous

what do you mean? it cut the x-axis at 2.2

- Loser66

but you can use it. now find f(0) =?

- Loser66

replace 0 into f(x) , what do you get?

- Loser66

|dw:1442536923437:dw|

- anonymous

-5

- Loser66

ok, so f(0) =-5 < 0 , right?
now f (2) =?

- anonymous

-3

- anonymous

So what does that mean

- Loser66

that mean from 0 to 2 you still not pass the x-axis, since f(x) <0 for both. Now test f(3)

- anonymous

22

- Loser66

yes, so from 2 to 3 , f(x) go from negative to positive, that is you have one zero on that interval, right?

- Loser66

Now, use synthetic,

- Loser66

forget it, short cut

- anonymous

okay haha

- Loser66

|dw:1442537533408:dw|

- Loser66

now, fill up

- anonymous

Again, the lesson didnt really explain this stuff so I'm not sure how to use that

- Loser66

you don't know synthetic?

- anonymous

not completely, looking it up now trying to learn real quick

- Loser66

|dw:1442537680182:dw|

- Loser66

|dw:1442537714109:dw|

- Loser66

now 4 *6 =?

- anonymous

24

- Loser66

|dw:1442537779766:dw| got it?

- Loser66

the last one, you do, the same

- anonymous

would it be 4*21 = 84 or 4*24 =96?

- Loser66

|dw:1442537857152:dw|

- Loser66

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.

- anonymous

oh okay i see

- anonymous

that's easy then haha

- Loser66

|dw:1442538058710:dw|

- Loser66

ok, now,
4 >0 , right?

- anonymous

yes

- Loser66

look at the bottom row, they are 2,6,21,79 right?

- anonymous

yes

- Loser66

and all them are positive, right?

- anonymous

yes

- Loser66

Hence 4 is upper bound for the zeros

- anonymous

woah i see, a lot goes into it

- anonymous

so back to the original question the answer would be no

- Loser66

yup

- anonymous

ok thanks for all of the help, what would happen if one of them was negative?

- Loser66

you test, if k <0, the bottom line is +, -,+,- alternatively, then k is lower bound

- Loser66

learn from it http://www.howe-two.com/mathematicat/bound.html

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