Question

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 = -1; f(x) = 4x3 - 2x2 + 2x + 4; Lower bound?

Answer 1

did you do the synthetic division?

Answer 2

@satellite73 no i dont know how to

Answer 3

it is hard for me to write it here but i can try if you did not do it

Answer 4

\[ f(x) = 4x^3 - 2x^2 + 2x + 4\] list the coefficients
4 -2 2 4

Answer 5

okay

Answer 6

put a \(-1\) on the side

4 -2 2 4
-1
______________

Answer 7

bring down the 4
4 -2 2 4
-1
______________
4

Answer 8

then \(-1\times 4=-4\)
4 -2 2 4
-1 -4
______________
4

Answer 9

add and get \(-6\)
4 -2 2 4
-1 -4
______________
4 -6

Answer 10

\(-1\times -6=6\)
4 -2 2 4
-1 -4 6
______________
4 -6

Answer 11

add and get \(8\)
4 -2 2 4
-1 -4 6
______________
4 -6 8

Answer 12

finally \(-1\times 8=-8\)
4 -2 2 4
-1 -4 6 -8
______________
4 -6 8

Answer 13

add and get \(-4\)
4 -2 2 4
-1 -4 6 -8
______________
4 -6 8 4

Answer 14

now look only at the bottom row
4 -6 8 -4

Answer 15

if the signs were all the same, it would be an upper bound, but they are not
however, since the signs alternate + - + -
then it IS a lower bound

Answer 16

not sure how much of that you followed, but that is the method for synthetic division

Answer 17

thanks so much i have another question if you dont mind helping @satellite73

Answer 18

go ahead and ask,
i will try

Answer 19

Divide f(x) by d(x), and write a summary statement in the form indicated.

f(x) = x4 + 5x3 - 2x2 + 5x - 3; d(x) = x2 + 1

Answer 20

it is almost impossible to write long division here but i guess we can try it

Answer 21

x^2 + 1 ( x^4 + 5x^3 -2x^2 + 5x -3

Answer 22

\(x^2\) goes in to \(x^4\) \(x^2\) times


x^2
x^2 + 1 ( x^4 + 5x^3 -2x^2 + 5x -3

Answer 23

then \(x^2\times (x^2+1)=x^4+x^2\)

x^2
x^2 + 1 ( x^4 + 5x^3 -2x^2 + 5x -3
x^4 +x^2

Answer 24

subtract and get

x^2
x^2 + 1 ( x^4 + 5x^3 -2x^2 + 5x -3
x^4 +x^2
-------------------------
5x^3 - 3x^2

Answer 25

ick this is a pain
\(x^2\) goes in to \(5x^3\) \(5x\) times

x^2 + 5x
x^2 + 1 ( x^4 + 5x^3 -2x^2 + 5x -3
x^4 +x^2
-------------------------
5x^3 - 3x^2

Answer 26

okay then how do you get the remainder

Answer 27

and \(5x(x^2+1)=5x^3+5x\)

x^2 + 5x
x^2 + 1 ( x^4 + 5x^3 -2x^2 + 5x -3
x^4 +x^2
-------------------------
5x^3 - 3x^2
5x^3 + 5x

Answer 28

gotta divide

Answer 29

subtract and get
x^2 + 5x
x^2 + 1 ( x^4 + 5x^3 -2x^2 + 5x -3
x^4 +x^2
-------------------------
5x^3 - 3x^2
5x^3 + 5x
-----------------------
-3x^2 + 0

Answer 30

f(x) = (x2 + 1)( x2 + 5x - 3)

f(x) = (x2 + 1)( x2 + 5x - 3) - 15x - 9

f(x) = (x2 + 1)( x2 - 5x - 3) - 15x - 9

f(x) = (x2 + 1)( x2 - 5x - 3)

These are my answer choices how do i get to this

Answer 31

well you could always multiply out and see which one works

Answer 32

or just cheat

Answer 33

how do you cheat

Answer 34

http://www.wolframalpha.com/input/?i=%28+x^4+%2B+5x^3+-+2x^2+%2B+5x+-+3%29%2F%28x^2+%2B+1%29

Answer 35

you see if you divide you get \(5x^2+5x-3\)

Answer 36

so no remainder