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Answer :
We begin with the polynomial
[tex]$$
f(x)=2x^5+17x^4+0x^3+6x^2+4x+39.
$$[/tex]
To check whether [tex]$3$[/tex] is an upper bound for the real zeros of [tex]$f(x)$[/tex], we use synthetic division with the candidate [tex]$3$[/tex]. List the coefficients in descending order:
[tex]$$
2,\ 17,\ 0,\ 6,\ 4,\ 39.
$$[/tex]
The synthetic division process is as follows:
1. Write down the candidate [tex]$3$[/tex] to the left and the coefficients in a row.
2. Bring the first coefficient, [tex]$2$[/tex], straight down.
3. Multiply the number just brought down by [tex]$3$[/tex]:
[tex]$$
2\times 3=6.
$$[/tex]
Add this to the next coefficient:
[tex]$$
17+6=23.
$$[/tex]
4. Multiply [tex]$23$[/tex] by [tex]$3$[/tex]:
[tex]$$
23\times 3=69.
$$[/tex]
Add to the next coefficient:
[tex]$$
0+69=69.
$$[/tex]
5. Multiply [tex]$69$[/tex] by [tex]$3$[/tex]:
[tex]$$
69\times 3=207.
$$[/tex]
Add to the next coefficient:
[tex]$$
6+207=213.
$$[/tex]
6. Multiply [tex]$213$[/tex] by [tex]$3$[/tex]:
[tex]$$
213\times 3=639.
$$[/tex]
Add to the next coefficient:
[tex]$$
4+639=643.
$$[/tex]
7. Multiply [tex]$643$[/tex] by [tex]$3$[/tex]:
[tex]$$
643\times 3=1929.
$$[/tex]
Add to the last coefficient:
[tex]$$
39+1929=1968.
$$[/tex]
The resulting synthetic division row is:
[tex]$$
[2,\ 23,\ 69,\ 213,\ 643,\ 1968].
$$[/tex]
Since every number in this row is non-negative, by the upper bound theorem, [tex]$3$[/tex] is an upper bound for the real zeros of [tex]$f(x)$[/tex].
Thus, the answer to part (a) is that [tex]$3$[/tex] is an upper bound for the real zeros of [tex]$f(x)$[/tex].
[tex]$$
f(x)=2x^5+17x^4+0x^3+6x^2+4x+39.
$$[/tex]
To check whether [tex]$3$[/tex] is an upper bound for the real zeros of [tex]$f(x)$[/tex], we use synthetic division with the candidate [tex]$3$[/tex]. List the coefficients in descending order:
[tex]$$
2,\ 17,\ 0,\ 6,\ 4,\ 39.
$$[/tex]
The synthetic division process is as follows:
1. Write down the candidate [tex]$3$[/tex] to the left and the coefficients in a row.
2. Bring the first coefficient, [tex]$2$[/tex], straight down.
3. Multiply the number just brought down by [tex]$3$[/tex]:
[tex]$$
2\times 3=6.
$$[/tex]
Add this to the next coefficient:
[tex]$$
17+6=23.
$$[/tex]
4. Multiply [tex]$23$[/tex] by [tex]$3$[/tex]:
[tex]$$
23\times 3=69.
$$[/tex]
Add to the next coefficient:
[tex]$$
0+69=69.
$$[/tex]
5. Multiply [tex]$69$[/tex] by [tex]$3$[/tex]:
[tex]$$
69\times 3=207.
$$[/tex]
Add to the next coefficient:
[tex]$$
6+207=213.
$$[/tex]
6. Multiply [tex]$213$[/tex] by [tex]$3$[/tex]:
[tex]$$
213\times 3=639.
$$[/tex]
Add to the next coefficient:
[tex]$$
4+639=643.
$$[/tex]
7. Multiply [tex]$643$[/tex] by [tex]$3$[/tex]:
[tex]$$
643\times 3=1929.
$$[/tex]
Add to the last coefficient:
[tex]$$
39+1929=1968.
$$[/tex]
The resulting synthetic division row is:
[tex]$$
[2,\ 23,\ 69,\ 213,\ 643,\ 1968].
$$[/tex]
Since every number in this row is non-negative, by the upper bound theorem, [tex]$3$[/tex] is an upper bound for the real zeros of [tex]$f(x)$[/tex].
Thus, the answer to part (a) is that [tex]$3$[/tex] is an upper bound for the real zeros of [tex]$f(x)$[/tex].
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Rewritten by : Jeany
