Answer :
We want to determine the possible numbers of positive real zeros (which we call [tex]\( p \)[/tex] zeros) and imaginary (non-real complex) zeros for the polynomial
[tex]$$
f(x) = 10x^5 - 5x^4 - 46x^3 + 23x^2 - 84x + 42.
$$[/tex]
Since the polynomial is of degree 5, it has exactly 5 zeros (counting multiplicities and including complex zeros).
──────────────────────────────
Step 1. Determine the possible number of positive real zeros
To do this, we use Descartes’ Rule of Signs. First, look at the coefficients of [tex]\( f(x) \)[/tex]:
[tex]$$
10,\quad -5,\quad -46,\quad 23,\quad -84,\quad 42.
$$[/tex]
Now, note the signs as follows:
- [tex]\( 10 \)[/tex] is [tex]\( + \)[/tex]
- [tex]\( -5 \)[/tex] is [tex]\( - \)[/tex]
- [tex]\( -46 \)[/tex] is [tex]\( - \)[/tex]
- [tex]\( 23 \)[/tex] is [tex]\( + \)[/tex]
- [tex]\( -84 \)[/tex] is [tex]\( - \)[/tex]
- [tex]\( 42 \)[/tex] is [tex]\( + \)[/tex]
Now, count the sign changes (ignoring any zeros):
1. [tex]\( + \)[/tex] to [tex]\( - \)[/tex] gives a change (1).
2. [tex]\( - \)[/tex] to [tex]\( - \)[/tex] gives no change.
3. [tex]\( - \)[/tex] to [tex]\( + \)[/tex] gives a change (2).
4. [tex]\( + \)[/tex] to [tex]\( - \)[/tex] gives a change (3).
5. [tex]\( - \)[/tex] to [tex]\( + \)[/tex] gives a change (4).
Thus, we have 4 sign changes. This means the number of positive real zeros is either 4, or less than 4 by an even number. Hence, the possible counts for positive zeros are
[tex]$$
4,\quad 2,\quad \text{or} \quad 0.
$$[/tex]
──────────────────────────────
Step 2. Determine the number of negative real zeros
Now, we replace [tex]\( x \)[/tex] by [tex]\( -x \)[/tex] to get [tex]\( f(-x) \)[/tex]:
[tex]\[
\begin{aligned}
f(-x) &= 10(-x)^5 - 5(-x)^4 - 46(-x)^3 + 23(-x)^2 - 84(-x) + 42 \\
&= 10(-x^5) - 5(x^4) - 46(-x^3) + 23(x^2) + 84x + 42 \\
&= -10x^5 - 5x^4 + 46x^3 + 23x^2 + 84x + 42.
\end{aligned}
\][/tex]
The coefficients of [tex]\( f(-x) \)[/tex] and their signs are:
- [tex]\( -10 \)[/tex] is [tex]\( - \)[/tex]
- [tex]\( -5 \)[/tex] is [tex]\( - \)[/tex]
- [tex]\( 46 \)[/tex] is [tex]\( + \)[/tex]
- [tex]\( 23 \)[/tex] is [tex]\( + \)[/tex]
- [tex]\( 84 \)[/tex] is [tex]\( + \)[/tex]
- [tex]\( 42 \)[/tex] is [tex]\( + \)[/tex]
Count the sign changes:
1. [tex]\( - \)[/tex] to [tex]\( - \)[/tex]: no change.
2. [tex]\( - \)[/tex] to [tex]\( + \)[/tex]: one change.
3. Remaining transitions are from [tex]\( + \)[/tex] to [tex]\( + \)[/tex] (no further change).
Thus, there is exactly 1 sign change. According to Descartes’ Rule, there is exactly 1 negative real zero.
──────────────────────────────
Step 3. Determine the possible numbers of imaginary zeros
Non-real complex (imaginary) zeros always come in conjugate pairs. Since the total number of zeros is 5, if we denote the number of positive zeros by [tex]\( p \)[/tex] (possible values [tex]\(4\)[/tex], [tex]\(2\)[/tex], or [tex]\(0\)[/tex]) and note there is 1 negative zero, then the number of real zeros is
[tex]$$
p + 1.
$$[/tex]
The number of imaginary (non-real) zeros will be
[tex]$$
5 - (p + 1) = 4 - p.
$$[/tex]
Now, consider the different cases:
1. If [tex]\( p = 4 \)[/tex]:
- Total real zeros [tex]\(= 4 + 1 = 5\)[/tex].
- Imaginary zeros [tex]\(= 5 - 5 = 0\)[/tex].
2. If [tex]\( p = 2 \)[/tex]:
- Total real zeros [tex]\(= 2 + 1 = 3\)[/tex].
- Imaginary zeros [tex]\(= 5 - 3 = 2\)[/tex].
3. If [tex]\( p = 0 \)[/tex]:
- Total real zeros [tex]\(= 0 + 1 = 1\)[/tex].
- Imaginary zeros [tex]\(= 5 - 1 = 4\)[/tex].
──────────────────────────────
Final Answer
The possible numbers for the positive ( [tex]\( p \)[/tex] ) zeros are
[tex]$$
4,\quad 2,\quad \text{or} \quad 0.
$$[/tex]
And the corresponding possible numbers of imaginary (non-real complex) zeros are
[tex]$$
0,\quad 2,\quad \text{or} \quad 4,
$$[/tex]
respectively.
Thus, the function has exactly 1 negative real zero, and the possibilities are:
- 4 positive real zeros, 1 negative real zero, and 0 imaginary zeros;
- 2 positive real zeros, 1 negative real zero, and 2 imaginary zeros;
- 0 positive real zeros, 1 negative real zero, and 4 imaginary zeros.
[tex]$$
f(x) = 10x^5 - 5x^4 - 46x^3 + 23x^2 - 84x + 42.
$$[/tex]
Since the polynomial is of degree 5, it has exactly 5 zeros (counting multiplicities and including complex zeros).
──────────────────────────────
Step 1. Determine the possible number of positive real zeros
To do this, we use Descartes’ Rule of Signs. First, look at the coefficients of [tex]\( f(x) \)[/tex]:
[tex]$$
10,\quad -5,\quad -46,\quad 23,\quad -84,\quad 42.
$$[/tex]
Now, note the signs as follows:
- [tex]\( 10 \)[/tex] is [tex]\( + \)[/tex]
- [tex]\( -5 \)[/tex] is [tex]\( - \)[/tex]
- [tex]\( -46 \)[/tex] is [tex]\( - \)[/tex]
- [tex]\( 23 \)[/tex] is [tex]\( + \)[/tex]
- [tex]\( -84 \)[/tex] is [tex]\( - \)[/tex]
- [tex]\( 42 \)[/tex] is [tex]\( + \)[/tex]
Now, count the sign changes (ignoring any zeros):
1. [tex]\( + \)[/tex] to [tex]\( - \)[/tex] gives a change (1).
2. [tex]\( - \)[/tex] to [tex]\( - \)[/tex] gives no change.
3. [tex]\( - \)[/tex] to [tex]\( + \)[/tex] gives a change (2).
4. [tex]\( + \)[/tex] to [tex]\( - \)[/tex] gives a change (3).
5. [tex]\( - \)[/tex] to [tex]\( + \)[/tex] gives a change (4).
Thus, we have 4 sign changes. This means the number of positive real zeros is either 4, or less than 4 by an even number. Hence, the possible counts for positive zeros are
[tex]$$
4,\quad 2,\quad \text{or} \quad 0.
$$[/tex]
──────────────────────────────
Step 2. Determine the number of negative real zeros
Now, we replace [tex]\( x \)[/tex] by [tex]\( -x \)[/tex] to get [tex]\( f(-x) \)[/tex]:
[tex]\[
\begin{aligned}
f(-x) &= 10(-x)^5 - 5(-x)^4 - 46(-x)^3 + 23(-x)^2 - 84(-x) + 42 \\
&= 10(-x^5) - 5(x^4) - 46(-x^3) + 23(x^2) + 84x + 42 \\
&= -10x^5 - 5x^4 + 46x^3 + 23x^2 + 84x + 42.
\end{aligned}
\][/tex]
The coefficients of [tex]\( f(-x) \)[/tex] and their signs are:
- [tex]\( -10 \)[/tex] is [tex]\( - \)[/tex]
- [tex]\( -5 \)[/tex] is [tex]\( - \)[/tex]
- [tex]\( 46 \)[/tex] is [tex]\( + \)[/tex]
- [tex]\( 23 \)[/tex] is [tex]\( + \)[/tex]
- [tex]\( 84 \)[/tex] is [tex]\( + \)[/tex]
- [tex]\( 42 \)[/tex] is [tex]\( + \)[/tex]
Count the sign changes:
1. [tex]\( - \)[/tex] to [tex]\( - \)[/tex]: no change.
2. [tex]\( - \)[/tex] to [tex]\( + \)[/tex]: one change.
3. Remaining transitions are from [tex]\( + \)[/tex] to [tex]\( + \)[/tex] (no further change).
Thus, there is exactly 1 sign change. According to Descartes’ Rule, there is exactly 1 negative real zero.
──────────────────────────────
Step 3. Determine the possible numbers of imaginary zeros
Non-real complex (imaginary) zeros always come in conjugate pairs. Since the total number of zeros is 5, if we denote the number of positive zeros by [tex]\( p \)[/tex] (possible values [tex]\(4\)[/tex], [tex]\(2\)[/tex], or [tex]\(0\)[/tex]) and note there is 1 negative zero, then the number of real zeros is
[tex]$$
p + 1.
$$[/tex]
The number of imaginary (non-real) zeros will be
[tex]$$
5 - (p + 1) = 4 - p.
$$[/tex]
Now, consider the different cases:
1. If [tex]\( p = 4 \)[/tex]:
- Total real zeros [tex]\(= 4 + 1 = 5\)[/tex].
- Imaginary zeros [tex]\(= 5 - 5 = 0\)[/tex].
2. If [tex]\( p = 2 \)[/tex]:
- Total real zeros [tex]\(= 2 + 1 = 3\)[/tex].
- Imaginary zeros [tex]\(= 5 - 3 = 2\)[/tex].
3. If [tex]\( p = 0 \)[/tex]:
- Total real zeros [tex]\(= 0 + 1 = 1\)[/tex].
- Imaginary zeros [tex]\(= 5 - 1 = 4\)[/tex].
──────────────────────────────
Final Answer
The possible numbers for the positive ( [tex]\( p \)[/tex] ) zeros are
[tex]$$
4,\quad 2,\quad \text{or} \quad 0.
$$[/tex]
And the corresponding possible numbers of imaginary (non-real complex) zeros are
[tex]$$
0,\quad 2,\quad \text{or} \quad 4,
$$[/tex]
respectively.
Thus, the function has exactly 1 negative real zero, and the possibilities are:
- 4 positive real zeros, 1 negative real zero, and 0 imaginary zeros;
- 2 positive real zeros, 1 negative real zero, and 2 imaginary zeros;
- 0 positive real zeros, 1 negative real zero, and 4 imaginary zeros.
