Precalculus Examples

$f (x) = 2 x^{3} - 4 x^{2} - 14 x + 28$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28$

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 4, \pm 7, \pm \frac{7}{2}, \pm 14, \pm 28$

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.

$2 {(2)}^{3} - 4 {(2)}^{2} - 14 \cdot 2 + 28$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 2$ is a root of the polynomial.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply $2$ by ${(2)}^{3}$ by adding the exponents.

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Step 4.1.1.1

Multiply $2$ by ${(2)}^{3}$ .

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Step 4.1.1.1.1

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

$2^{1} {(2)}^{3} - 4 {(2)}^{2} - 14 \cdot 2 + 28$

Step 4.1.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2^{1 + 3} - 4 {(2)}^{2} - 14 \cdot 2 + 28$

Step 4.1.1.2

Add $1$ and $3$ .

$2^{4} - 4 {(2)}^{2} - 14 \cdot 2 + 28$

Step 4.1.2

Raise $2$ to the power of $4$ .

$16 - 4 {(2)}^{2} - 14 \cdot 2 + 28$

Step 4.1.3

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

$16 - 4 \cdot 4 - 14 \cdot 2 + 28$

Step 4.1.4

Multiply $- 4$ by $4$ .

$16 - 16 - 14 \cdot 2 + 28$

Step 4.1.5

Multiply $- 14$ by $2$ .

$16 - 16 - 28 + 28$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $16$ from $16$ .

$0 - 28 + 28$

Step 4.2.2

Subtract $28$ from $0$ .

$- 28 + 28$

Step 4.2.3

Add $- 28$ and $28$ .

$0$

Step 5

Since $2$ is a known root, divide the polynomial by $x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{3} - 4 x^{2} - 14 x + 28}{x - 2}$

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.

$2$	$2$	$- 4$	$- 14$	$28$

Step 6.2

The first number in the dividend $(2)$ is put into the first position of the result area (below the horizontal line).

$2$	$2$	$- 4$	$- 14$	$28$

	$2$

Step 6.3

Multiply the newest entry in the result $(2)$ by the divisor $(2)$ and place the result of $(4)$ under the next term in the dividend $(- 4)$ .

$2$	$2$	$- 4$	$- 14$	$28$
		$4$
	$2$

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.

$2$	$2$	$- 4$	$- 14$	$28$
		$4$
	$2$	$0$

Step 6.5

Multiply the newest entry in the result $(0)$ by the divisor $(2)$ and place the result of $(0)$ under the next term in the dividend $(- 14)$ .

$2$	$2$	$- 4$	$- 14$	$28$
		$4$	$0$
	$2$	$0$

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.

$2$	$2$	$- 4$	$- 14$	$28$
		$4$	$0$
	$2$	$0$	$- 14$

Step 6.7

Multiply the newest entry in the result $(- 14)$ by the divisor $(2)$ and place the result of $(- 28)$ under the next term in the dividend $(28)$ .

$2$	$2$	$- 4$	$- 14$	$28$
		$4$	$0$	$- 28$
	$2$	$0$	$- 14$

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.

$2$	$2$	$- 4$	$- 14$	$28$
		$4$	$0$	$- 28$
	$2$	$0$	$- 14$	$0$

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.

$2 x^{2} + 0 x - 14$

Step 6.10

Simplify the quotient polynomial.

$2 x^{2} - 14$

Step 7

Factor $2$ out of $2 x^{2} - 14$ .

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Step 7.1

Factor $2$ out of $2 x^{2}$ .

$(x - 2) (2 (x^{2}) - 14)$

Step 7.2

Factor $2$ out of $- 14$ .

$(x - 2) (2 x^{2} + 2 \cdot - 7)$

Step 7.3

Factor $2$ out of $2 x^{2} + 2 \cdot - 7$ .

$(x - 2) (2 (x^{2} - 7))$

Step 8

Factor the left side of the equation.

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Step 8.1

Factor $2$ out of $2 x^{3} - 4 x^{2} - 14 x + 28$ .

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Step 8.1.1

Factor $2$ out of $2 x^{3}$ .

$2 (x^{3}) - 4 x^{2} - 14 x + 28 = 0$

Step 8.1.2

Factor $2$ out of $- 4 x^{2}$ .

$2 (x^{3}) + 2 (- 2 x^{2}) - 14 x + 28 = 0$

Step 8.1.3

Factor $2$ out of $- 14 x$ .

$2 (x^{3}) + 2 (- 2 x^{2}) + 2 (- 7 x) + 28 = 0$

Step 8.1.4

Factor $2$ out of $28$ .

$2 x^{3} + 2 (- 2 x^{2}) + 2 (- 7 x) + 2 \cdot 14 = 0$

Step 8.1.5

Factor $2$ out of $2 x^{3} + 2 (- 2 x^{2})$ .

$2 (x^{3} - 2 x^{2}) + 2 (- 7 x) + 2 \cdot 14 = 0$

Step 8.1.6

Factor $2$ out of $2 (x^{3} - 2 x^{2}) + 2 (- 7 x)$ .

$2 (x^{3} - 2 x^{2} - 7 x) + 2 \cdot 14 = 0$

Step 8.1.7

Factor $2$ out of $2 (x^{3} - 2 x^{2} - 7 x) + 2 \cdot 14$ .

$2 (x^{3} - 2 x^{2} - 7 x + 14) = 0$

Step 8.2

Factor out the greatest common factor from each group.

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Step 8.2.1

Group the first two terms and the last two terms.

$2 ((x^{3} - 2 x^{2}) - 7 x + 14) = 0$

Step 8.2.2

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

$2 (x^{2} (x - 2) - 7 (x - 2)) = 0$

Step 8.3

Factor.

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Step 8.3.1

Factor the polynomial by factoring out the greatest common factor, $x - 2$ .

$2 ((x - 2) (x^{2} - 7)) = 0$

Step 8.3.2

Remove unnecessary parentheses.

$2 (x - 2) (x^{2} - 7) = 0$

Step 9

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

$x - 2 = 0$

$x^{2} - 7 = 0$

Step 10

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 10.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 10.2

Add $2$ to both sides of the equation.

$x = 2$

Step 11

Set $x^{2} - 7$ equal to $0$ and solve for $x$ .

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Step 11.1

Set $x^{2} - 7$ equal to $0$ .

$x^{2} - 7 = 0$

Step 11.2

Solve $x^{2} - 7 = 0$ for $x$ .

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Step 11.2.1

Add $7$ to both sides of the equation.

$x^{2} = 7$

Step 11.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{7}$

Step 11.2.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 11.2.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{7}$

Step 11.2.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{7}$

Step 11.2.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{7}, - \sqrt{7}$

Step 12

The final solution is all the values that make $2 (x - 2) (x^{2} - 7) = 0$ true.

$x = 2, \sqrt{7}, - \sqrt{7}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$x = 2, \sqrt{7}, - \sqrt{7}$

Decimal Form:

$x = 2, 2.64575131 \dots, - 2.64575131 \dots$

Step 14