Precalculus Examples

$x^{3} + 3 x^{2} - 16 x - 48$

Step 1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 48$

$q = \pm 1$

Step 2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 48$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

${(- 3)}^{3} + 3 {(- 3)}^{2} - 16 \cdot - 3 - 48$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 3$ is a root of the polynomial.

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Step 4.1

Simplify each term.

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Step 4.1.1

Raise $- 3$ to the power of $3$ .

$- 27 + 3 {(- 3)}^{2} - 16 \cdot - 3 - 48$

Step 4.1.2

Raise $- 3$ to the power of $2$ .

$- 27 + 3 \cdot 9 - 16 \cdot - 3 - 48$

Step 4.1.3

Multiply $3$ by $9$ .

$- 27 + 27 - 16 \cdot - 3 - 48$

Step 4.1.4

Multiply $- 16$ by $- 3$ .

$- 27 + 27 + 48 - 48$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Add $- 27$ and $27$ .

$0 + 48 - 48$

Step 4.2.2

Add $0$ and $48$ .

$48 - 48$

Step 4.2.3

Subtract $48$ from $48$ .

$0$

Step 5

Since $- 3$ is a known root, divide the polynomial by $x + 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} + 3 x^{2} - 16 x - 48}{x + 3}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

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Step 6.1

Place the numbers representing the divisor and the dividend into a division-like configuration.

$- 3$	$1$	$3$	$- 16$	$- 48$

Step 6.2

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$- 3$	$1$	$3$	$- 16$	$- 48$

	$1$

Step 6.3

Multiply the newest entry in the result $(1)$ by the divisor $(- 3)$ and place the result of $(- 3)$ under the next term in the dividend $(3)$ .

$- 3$	$1$	$3$	$- 16$	$- 48$
		$- 3$
	$1$

Step 6.4

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 3$	$1$	$3$	$- 16$	$- 48$
		$- 3$
	$1$	$0$

Step 6.5

Multiply the newest entry in the result $(0)$ by the divisor $(- 3)$ and place the result of $(0)$ under the next term in the dividend $(- 16)$ .

$- 3$	$1$	$3$	$- 16$	$- 48$
		$- 3$	$0$
	$1$	$0$

Step 6.6

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 3$	$1$	$3$	$- 16$	$- 48$
		$- 3$	$0$
	$1$	$0$	$- 16$

Step 6.7

Multiply the newest entry in the result $(- 16)$ by the divisor $(- 3)$ and place the result of $(48)$ under the next term in the dividend $(- 48)$ .

$- 3$	$1$	$3$	$- 16$	$- 48$
		$- 3$	$0$	$48$
	$1$	$0$	$- 16$

Step 6.8

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 3$	$1$	$3$	$- 16$	$- 48$
		$- 3$	$0$	$48$
	$1$	$0$	$- 16$	$0$

Step 6.9

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$1 x^{2} + 0 x - 16$

Step 6.10

Simplify the quotient polynomial.

$x^{2} - 16$

Step 7

Rewrite $16$ as $4^{2}$ .

$(x + 3) x^{2} - 4^{2}$

Step 8

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

$(x + 3) (x + 4) (x - 4)$

Step 9

Factor the left side of the equation.

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Step 9.1

Factor out the greatest common factor from each group.

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Step 9.1.1

Group the first two terms and the last two terms.

$(x^{3} + 3 x^{2}) - 16 x - 48 = 0$

Step 9.1.2

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x + 3) - 16 (x + 3) = 0$

Step 9.2

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

$(x + 3) (x^{2} - 16) = 0$

Step 9.3

Rewrite $16$ as $4^{2}$ .

$(x + 3) (x^{2} - 4^{2}) = 0$

Step 9.4

Factor.

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Step 9.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

$(x + 3) ((x + 4) (x - 4)) = 0$

Step 9.4.2

Remove unnecessary parentheses.

$(x + 3) (x + 4) (x - 4) = 0$

Step 10

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 3 = 0$

$x + 4 = 0$

$x - 4 = 0$

Step 11

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 11.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 11.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 12

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 12.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 12.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 13

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 13.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 13.2

Add $4$ to both sides of the equation.

$x = 4$

Step 14

The final solution is all the values that make $(x + 3) (x + 4) (x - 4) = 0$ true.

$x = - 3, - 4, 4$

Step 15