Question Find the number of positive and negative real zeros of $g(x)=x^{3}+5x^{2}+9x-8$ .

Studdy Solution

STEP 1

Assumptions
1. The function given is $g(x) = x^{3} + 5x^{2} + 9x - 8$ .
2. We are to determine the possible number of positive and negative real zeros.
3. We will use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros.

STEP 2

Descartes' Rule of Signs states that the number of positive real zeros of a polynomial function $f(x)$ is either equal to the number of sign changes between consecutive coefficients of $f(x)$ , or less than that by an even number.

STEP 3

First, we will count the number of sign changes in the polynomial $g(x)$ to determine the possible number of positive real zeros.

STEP 4

The coefficients of $g(x)$ are 1, 5, 9, and -8. There is only one sign change (from 9 to -8).

STEP 5

According to Descartes' Rule of Signs, the possible number of positive real zeros is equal to the number of sign changes or less than that by an even number.

STEP 6

Since there is only one sign change, the possible number of positive real zeros can only be 1 (since it cannot be less than 1 by an even number).

STEP 7

To find the possible number of negative real zeros, we will apply Descartes' Rule of Signs to $g(-x)$ .

STEP 8

Substitute $-x$ for $x$ in $g(x)$ to get $g(-x)$ .
$g(-x) = (-x)^{3} + 5(-x)^{2} + 9(-x) - 8$

STEP 9

Simplify $g(-x)$ .
$g(-x) = -x^{3} + 5x^{2} - 9x - 8$

STEP 10

Now, count the number of sign changes in the polynomial $g(-x)$ .

STEP 11

The coefficients of $g(-x)$ are -1, 5, -9, and -8. There are two sign changes (from -1 to 5 and from -9 to -8).

STEP 12

According to Descartes' Rule of Signs, the possible number of negative real zeros is equal to the number of sign changes or less than that by an even number.

STEP 13

Since there are two sign changes, the possible number of negative real zeros can be 2 or 0 (2 minus an even number).

STEP 14

Now we can state the possible number of positive and negative real zeros for $g(x)$ .
(a) Possible number of positive real zeros: 1 (b) Possible number of negative real zeros: 2 or 0

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Question Find the number of positive and negative real zeros of $g(x)=x^{3}+5x^{2}+9x-8$ .

Studdy Solution

STEP 1

Assumptions
1. The function given is $g(x) = x^{3} + 5x^{2} + 9x - 8$ .
2. We are to determine the possible number of positive and negative real zeros.
3. We will use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros.

STEP 2

Descartes' Rule of Signs states that the number of positive real zeros of a polynomial function $f(x)$ is either equal to the number of sign changes between consecutive coefficients of $f(x)$ , or less than that by an even number.

STEP 3

First, we will count the number of sign changes in the polynomial $g(x)$ to determine the possible number of positive real zeros.

STEP 4

The coefficients of $g(x)$ are 1, 5, 9, and -8. There is only one sign change (from 9 to -8).

STEP 5

According to Descartes' Rule of Signs, the possible number of positive real zeros is equal to the number of sign changes or less than that by an even number.

STEP 6

Since there is only one sign change, the possible number of positive real zeros can only be 1 (since it cannot be less than 1 by an even number).

STEP 7

To find the possible number of negative real zeros, we will apply Descartes' Rule of Signs to $g(-x)$ .

STEP 8

Substitute $-x$ for $x$ in $g(x)$ to get $g(-x)$ .
$g(-x) = (-x)^{3} + 5(-x)^{2} + 9(-x) - 8$

STEP 9

Simplify $g(-x)$ .
$g(-x) = -x^{3} + 5x^{2} - 9x - 8$

STEP 10

Now, count the number of sign changes in the polynomial $g(-x)$ .

STEP 11

The coefficients of $g(-x)$ are -1, 5, -9, and -8. There are two sign changes (from -1 to 5 and from -9 to -8).

STEP 12

According to Descartes' Rule of Signs, the possible number of negative real zeros is equal to the number of sign changes or less than that by an even number.

STEP 13

Since there are two sign changes, the possible number of negative real zeros can be 2 or 0 (2 minus an even number).

STEP 14

Now we can state the possible number of positive and negative real zeros for $g(x)$ .
(a) Possible number of positive real zeros: 1 (b) Possible number of negative real zeros: 2 or 0

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Question Find the number of positive and negative real zeros of g(x)=x3+5x2+9x−8g(x)=x^{3}+5x^{2}+9x-8g(x)=x3+5x2+9x−8.

Studdy Solution

STEP 1

Assumptions1. The function given is g(x)=x3+5x2+9x−8 g(x) = x^{3} + 5x^{2} + 9x - 8 g(x)=x3+5x2+9x−8.2. We are to determine the possible number of positive and negative real zeros.3. We will use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros.

STEP 2

Descartes' Rule of Signs states that the number of positive real zeros of a polynomial function f(x) f(x) f(x) is either equal to the number of sign changes between consecutive coefficients of f(x) f(x) f(x), or less than that by an even number.

STEP 3

First, we will count the number of sign changes in the polynomial g(x) g(x) g(x) to determine the possible number of positive real zeros.

STEP 4

The coefficients of g(x) g(x) g(x) are 1, 5, 9, and -8. There is only one sign change (from 9 to -8).

STEP 5

According to Descartes' Rule of Signs, the possible number of positive real zeros is equal to the number of sign changes or less than that by an even number.

STEP 6

Since there is only one sign change, the possible number of positive real zeros can only be 1 (since it cannot be less than 1 by an even number).

STEP 7

To find the possible number of negative real zeros, we will apply Descartes' Rule of Signs to g(−x) g(-x) g(−x).

STEP 8

Substitute −x -x −x for x x x in g(x) g(x) g(x) to get g(−x) g(-x) g(−x).g(−x)=(−x)3+5(−x)2+9(−x)−8 g(-x) = (-x)^{3} + 5(-x)^{2} + 9(-x) - 8 g(−x)=(−x)3+5(−x)2+9(−x)−8

STEP 9

Simplify g(−x) g(-x) g(−x).g(−x)=−x3+5x2−9x−8 g(-x) = -x^{3} + 5x^{2} - 9x - 8 g(−x)=−x3+5x2−9x−8

STEP 10

Now, count the number of sign changes in the polynomial g(−x) g(-x) g(−x).

STEP 11

The coefficients of g(−x) g(-x) g(−x) are -1, 5, -9, and -8. There are two sign changes (from -1 to 5 and from -9 to -8).

STEP 12

According to Descartes' Rule of Signs, the possible number of negative real zeros is equal to the number of sign changes or less than that by an even number.

STEP 13

Since there are two sign changes, the possible number of negative real zeros can be 2 or 0 (2 minus an even number).

STEP 14

Now we can state the possible number of positive and negative real zeros for g(x) g(x) g(x).(a) Possible number of positive real zeros: 1 (b) Possible number of negative real zeros: 2 or 0

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Question Find the number of positive and negative real zeros of $g(x)=x^{3}+5x^{2}+9x-8$ .

Assumptions
1. The function given is $g(x) = x^{3} + 5x^{2} + 9x - 8$ .
2. We are to determine the possible number of positive and negative real zeros.
3. We will use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros.

Descartes' Rule of Signs states that the number of positive real zeros of a polynomial function $f(x)$ is either equal to the number of sign changes between consecutive coefficients of $f(x)$ , or less than that by an even number.

First, we will count the number of sign changes in the polynomial $g(x)$ to determine the possible number of positive real zeros.

The coefficients of $g(x)$ are 1, 5, 9, and -8. There is only one sign change (from 9 to -8).

To find the possible number of negative real zeros, we will apply Descartes' Rule of Signs to $g(-x)$ .

Substitute $-x$ for $x$ in $g(x)$ to get $g(-x)$ .
$g(-x) = (-x)^{3} + 5(-x)^{2} + 9(-x) - 8$

Simplify $g(-x)$ .
$g(-x) = -x^{3} + 5x^{2} - 9x - 8$

Now, count the number of sign changes in the polynomial $g(-x)$ .

The coefficients of $g(-x)$ are -1, 5, -9, and -8. There are two sign changes (from -1 to 5 and from -9 to -8).

Now we can state the possible number of positive and negative real zeros for $g(x)$ .
(a) Possible number of positive real zeros: 1 (b) Possible number of negative real zeros: 2 or 0