Finite Math Examples

x^{2} - 2 x - 3

$x^{2} - 2 x - 3$

Step 1

To find the possible number of positive roots, look at the signs on the coefficients and count the number of times the signs on the coefficients change from positive to negative or negative to positive.

f (x) = x^{2} - 2 x - 3

$f (x) = x^{2} - 2 x - 3$

Step 2

Since there is

1

$1$ sign change from the highest order term to the lowest, there is at most

1

$1$ positive root (Descartes' Rule of Signs).

Positive Roots:

1

$1$

Step 3

To find the possible number of negative roots, replace

x

$x$ with

- x

$- x$ and repeat the sign comparison.

f (- x) = {(- x)}^{2} - 2 (- x) - 3

$f (- x) = {(- x)}^{2} - 2 (- x) - 3$

Step 4

Simplify each term.

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Step 4.1

Apply the product rule to

- x

$- x$ .

f (- x) = {(- 1)}^{2} x^{2} - 2 (- x) - 3

$f (- x) = {(- 1)}^{2} x^{2} - 2 (- x) - 3$

Step 4.2

Raise

- 1

$- 1$ to the power of

2

$2$ .

f (- x) = 1 x^{2} - 2 (- x) - 3

$f (- x) = 1 x^{2} - 2 (- x) - 3$

Step 4.3

Multiply

x^{2}

$x^{2}$ by

1

$1$ .

f (- x) = x^{2} - 2 (- x) - 3

$f (- x) = x^{2} - 2 (- x) - 3$

Step 4.4

Multiply

- 1

$- 1$ by

- 2

$- 2$ .

f (- x) = x^{2} + 2 x - 3

$f (- x) = x^{2} + 2 x - 3$

f (- x) = x^{2} + 2 x - 3

$f (- x) = x^{2} + 2 x - 3$

Step 5

Since there is

1

$1$ sign change from the highest order term to the lowest, there is at most

1

$1$ negative root (Descartes' Rule of Signs).

Negative Roots:

1

$1$

Step 6

The possible number of positive roots is

1

$1$ , and the possible number of negative roots is

1

$1$ .

Positive Roots:

1

$1$

Negative Roots:

1

$1$

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