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Answer :
We start with the polynomial
[tex]$$
f(x)=2x^5+17x^4+0x^3+6x^2+4x+39.
$$[/tex]
To see if [tex]$3$[/tex] is an upper bound for the real zeros of [tex]$f(x)$[/tex], we perform synthetic division with the candidate [tex]$3$[/tex]. The process is as follows:
1. Write down the coefficients:
[tex]$$
2,\; 17,\; 0,\; 6,\; 4,\; 39.
$$[/tex]
2. Bring down the first coefficient:
[tex]$$
2.
$$[/tex]
3. Multiply this number by [tex]$3$[/tex], then add the result to the next coefficient:
- Multiply: [tex]$2 \times 3 = 6$[/tex].
- Add: [tex]$17 + 6 = 23$[/tex].
4. Multiply the new value by [tex]$3$[/tex], then add to the next coefficient:
- Multiply: [tex]$23 \times 3 = 69$[/tex].
- Add: [tex]$0 + 69 = 69$[/tex].
5. Continue this process:
- Multiply: [tex]$69 \times 3 = 207$[/tex].
- Add: [tex]$6 + 207 = 213$[/tex].
6. Next, multiply [tex]$213$[/tex] by [tex]$3$[/tex]:
- Multiply: [tex]$213 \times 3 = 639$[/tex].
- Add: [tex]$4 + 639 = 643$[/tex].
7. Finally, multiply [tex]$643$[/tex] by [tex]$3$[/tex]:
- Multiply: [tex]$643 \times 3 = 1929$[/tex].
- Add: [tex]$39 + 1929 = 1968$[/tex].
The final list of numbers obtained in the synthetic division is
[tex]$$
2,\; 23,\; 69,\; 213,\; 643,\; 1968.
$$[/tex]
Since all the numbers in this list are nonnegative, the upper bound theorem tells us that [tex]$3$[/tex] is indeed an upper bound for the real zeros of [tex]$f(x)$[/tex].
Thus, the answer to part (a) is:
[tex]$$
\textbf{Yes, 3 is an upper bound for the real zeros of } f(x).
$$[/tex]
[tex]$$
f(x)=2x^5+17x^4+0x^3+6x^2+4x+39.
$$[/tex]
To see if [tex]$3$[/tex] is an upper bound for the real zeros of [tex]$f(x)$[/tex], we perform synthetic division with the candidate [tex]$3$[/tex]. The process is as follows:
1. Write down the coefficients:
[tex]$$
2,\; 17,\; 0,\; 6,\; 4,\; 39.
$$[/tex]
2. Bring down the first coefficient:
[tex]$$
2.
$$[/tex]
3. Multiply this number by [tex]$3$[/tex], then add the result to the next coefficient:
- Multiply: [tex]$2 \times 3 = 6$[/tex].
- Add: [tex]$17 + 6 = 23$[/tex].
4. Multiply the new value by [tex]$3$[/tex], then add to the next coefficient:
- Multiply: [tex]$23 \times 3 = 69$[/tex].
- Add: [tex]$0 + 69 = 69$[/tex].
5. Continue this process:
- Multiply: [tex]$69 \times 3 = 207$[/tex].
- Add: [tex]$6 + 207 = 213$[/tex].
6. Next, multiply [tex]$213$[/tex] by [tex]$3$[/tex]:
- Multiply: [tex]$213 \times 3 = 639$[/tex].
- Add: [tex]$4 + 639 = 643$[/tex].
7. Finally, multiply [tex]$643$[/tex] by [tex]$3$[/tex]:
- Multiply: [tex]$643 \times 3 = 1929$[/tex].
- Add: [tex]$39 + 1929 = 1968$[/tex].
The final list of numbers obtained in the synthetic division is
[tex]$$
2,\; 23,\; 69,\; 213,\; 643,\; 1968.
$$[/tex]
Since all the numbers in this list are nonnegative, the upper bound theorem tells us that [tex]$3$[/tex] is indeed an upper bound for the real zeros of [tex]$f(x)$[/tex].
Thus, the answer to part (a) is:
[tex]$$
\textbf{Yes, 3 is an upper bound for the real zeros of } f(x).
$$[/tex]
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Rewritten by : Jeany
