Precalculus Examples

$x^{3} + 5 x$

Step 1

Factor out the GCF of $x$ from $x^{3} + 5 x$ .

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Step 1.1

Factor out the GCF of $x$ from each term in the polynomial.

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Step 1.1.1

Factor out the GCF of $x$ from the expression $x^{3}$ .

$x (x^{2}) + 5 x$

Step 1.1.2

Factor out the GCF of $x$ from the expression $5 x$ .

$x (x^{2}) + x (5)$

Step 1.2

Since all the terms share a common factor of $x$ , it can be factored out of each term.

$x (x^{2} + 5)$

Step 2

Apply Descartes' rule on the inside expression $x^{2} + 5$ .

$x^{2} + 5$

Step 3

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.

$f (x) = x^{2} + 5$

Step 4

Since there are $0$ sign changes from the highest order term to the lowest, there are at most $0$ positive roots (Descartes' Rule of Signs).

Positive Roots: $0$

Step 5

To find the possible number of negative roots, replace $x$ with $- x$ and repeat the sign comparison.

$f (- x) = {(- x)}^{2} + 5$

Step 6

Simplify each term.

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Step 6.1

Apply the product rule to $- x$ .

$f (- x) = {(- 1)}^{2} x^{2} + 5$

Step 6.2

Raise $- 1$ to the power of $2$ .

$f (- x) = 1 x^{2} + 5$

Step 6.3

Multiply $x^{2}$ by $1$ .

$f (- x) = x^{2} + 5$

Step 7

Since there are $0$ sign changes from the highest order term to the lowest, there are at most $0$ negative roots (Descartes' Rule of Signs).

Negative Roots: $0$

Step 8

The possible number of positive roots is $0$ , and the possible number of negative roots is $0$ .

Positive Roots: $0$

Negative Roots: $0$