Precalculus Examples

$9 x^{5} - 27 x^{4} - 10 x^{3} + 30 x^{2} + x - 3$

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.

$p = \pm 1, \pm 3$

$q = \pm 1, \pm 3, \pm 9$

Step 2

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

$\pm 1, \pm \frac{1}{3}, \pm \frac{1}{9}, \pm 3$

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.

$9 {(1)}^{5} - 27 {(1)}^{4} - 10 {(1)}^{3} + 30 {(1)}^{2} + 1 - 3$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 1$ is a root of the polynomial.

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Step 4.1

Remove parentheses.

$9 {(1)}^{5} - 27 {(1)}^{4} - 10 {(1)}^{3} + 30 {(1)}^{2} + 1 - 3$

Step 4.2

Simplify each term.

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Step 4.2.1

One to any power is one.

$9 \cdot 1 - 27 {(1)}^{4} - 10 {(1)}^{3} + 30 {(1)}^{2} + 1 - 3$

Step 4.2.2

Multiply $9$ by $1$ .

$9 - 27 {(1)}^{4} - 10 {(1)}^{3} + 30 {(1)}^{2} + 1 - 3$

Step 4.2.3

One to any power is one.

$9 - 27 \cdot 1 - 10 {(1)}^{3} + 30 {(1)}^{2} + 1 - 3$

Step 4.2.4

Multiply $- 27$ by $1$ .

$9 - 27 - 10 {(1)}^{3} + 30 {(1)}^{2} + 1 - 3$

Step 4.2.5

One to any power is one.

$9 - 27 - 10 \cdot 1 + 30 {(1)}^{2} + 1 - 3$

Step 4.2.6

Multiply $- 10$ by $1$ .

$9 - 27 - 10 + 30 {(1)}^{2} + 1 - 3$

Step 4.2.7

One to any power is one.

$9 - 27 - 10 + 30 \cdot 1 + 1 - 3$

Step 4.2.8

Multiply $30$ by $1$ .

$9 - 27 - 10 + 30 + 1 - 3$

$9 - 27 - 10 + 30 + 1 - 3$

Step 4.3

Simplify by adding and subtracting.

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Step 4.3.1

Subtract $27$ from $9$ .

$- 18 - 10 + 30 + 1 - 3$

Step 4.3.2

Subtract $10$ from $- 18$ .

$- 28 + 30 + 1 - 3$

Step 4.3.3

Add $- 28$ and $30$ .

$2 + 1 - 3$

Step 4.3.4

Add $2$ and $1$ .

$3 - 3$

Step 4.3.5

Subtract $3$ from $3$ .

$0$

$0$

$0$

Step 5

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{9 x^{5} - 27 x^{4} - 10 x^{3} + 30 x^{2} + x - 3}{x - 1}$

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.

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$

Step 6.2

The first number in the dividend $(9)$ is put into the first position of the result area (below the horizontal line).

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$

	$9$

Step 6.3

Multiply the newest entry in the result $(9)$ by the divisor $(1)$ and place the result of $(9)$ under the next term in the dividend $(- 27)$ .

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$
		$9$
	$9$

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.

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$
		$9$
	$9$	$- 18$

Step 6.5

Multiply the newest entry in the result $(- 18)$ by the divisor $(1)$ and place the result of $(- 18)$ under the next term in the dividend $(- 10)$ .

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$
		$9$	$- 18$
	$9$	$- 18$

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.

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$
		$9$	$- 18$
	$9$	$- 18$	$- 28$

Step 6.7

Multiply the newest entry in the result $(- 28)$ by the divisor $(1)$ and place the result of $(- 28)$ under the next term in the dividend $(30)$ .

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$
		$9$	$- 18$	$- 28$
	$9$	$- 18$	$- 28$

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.

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$
		$9$	$- 18$	$- 28$
	$9$	$- 18$	$- 28$	$2$

Step 6.9

Multiply the newest entry in the result $(2)$ by the divisor $(1)$ and place the result of $(2)$ under the next term in the dividend $(1)$ .

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$
		$9$	$- 18$	$- 28$	$2$
	$9$	$- 18$	$- 28$	$2$

Step 6.10

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

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$
		$9$	$- 18$	$- 28$	$2$
	$9$	$- 18$	$- 28$	$2$	$3$

Step 6.11

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

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$
		$9$	$- 18$	$- 28$	$2$	$3$
	$9$	$- 18$	$- 28$	$2$	$3$

Step 6.12

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

$1$	$9$	$- 27$	$- 10$	$30$	$1$	$- 3$
		$9$	$- 18$	$- 28$	$2$	$3$
	$9$	$- 18$	$- 28$	$2$	$3$	$0$

Step 6.13

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

$9 x^{4} + - 18 x^{3} - 28 x^{2} + 2 x + 3$

Step 6.14

Simplify the quotient polynomial.

$9 x^{4} - 18 x^{3} - 28 x^{2} + 2 x + 3$

$9 x^{4} - 18 x^{3} - 28 x^{2} + 2 x + 3$

Step 7

Solve the equation to find any remaining roots.

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Step 7.1

Factor the left side of the equation.

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Step 7.1.1

Regroup terms.

$- 18 x^{3} + 2 x + 9 x^{4} - 28 x^{2} + 3 = 0$

Step 7.1.2

Factor $- 2 x$ out of $- 18 x^{3} + 2 x$ .

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Step 7.1.2.1

Factor $- 2 x$ out of $- 18 x^{3}$ .

$- 2 x (9 x^{2}) + 2 x + 9 x^{4} - 28 x^{2} + 3 = 0$

Step 7.1.2.2

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

$- 2 x (9 x^{2}) - 2 x \cdot - 1 + 9 x^{4} - 28 x^{2} + 3 = 0$

Step 7.1.2.3

Factor $- 2 x$ out of $- 2 x (9 x^{2}) - 2 x (- 1)$ .

$- 2 x (9 x^{2} - 1) + 9 x^{4} - 28 x^{2} + 3 = 0$

$- 2 x (9 x^{2} - 1) + 9 x^{4} - 28 x^{2} + 3 = 0$

Step 7.1.3

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$- 2 x ({(3 x)}^{2} - 1) + 9 x^{4} - 28 x^{2} + 3 = 0$

Step 7.1.4

Rewrite $1$ as $1^{2}$ .

$- 2 x ({(3 x)}^{2} - 1^{2}) + 9 x^{4} - 28 x^{2} + 3 = 0$

Step 7.1.5

Factor.

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Step 7.1.5.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 1$ .

$- 2 x ((3 x + 1) (3 x - 1)) + 9 x^{4} - 28 x^{2} + 3 = 0$

Step 7.1.5.2

Remove unnecessary parentheses.

$- 2 x (3 x + 1) (3 x - 1) + 9 x^{4} - 28 x^{2} + 3 = 0$

$- 2 x (3 x + 1) (3 x - 1) + 9 x^{4} - 28 x^{2} + 3 = 0$

Step 7.1.6

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$- 2 x (3 x + 1) (3 x - 1) + 9 {(x^{2})}^{2} - 28 x^{2} + 3 = 0$

Step 7.1.7

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- 2 x (3 x + 1) (3 x - 1) + 9 u^{2} - 28 u + 3 = 0$

Step 7.1.8

Factor by grouping.

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Step 7.1.8.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 9 \cdot 3 = 27$ and whose sum is $b = - 28$ .

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Step 7.1.8.1.1

Factor $- 28$ out of $- 28 u$ .

$- 2 x (3 x + 1) (3 x - 1) + 9 u^{2} - 28 u + 3 = 0$

Step 7.1.8.1.2

Rewrite $- 28$ as $- 1$ plus $- 27$

$- 2 x (3 x + 1) (3 x - 1) + 9 u^{2} + (- 1 - 27) u + 3 = 0$

Step 7.1.8.1.3

Apply the distributive property.

$- 2 x (3 x + 1) (3 x - 1) + 9 u^{2} - 1 u - 27 u + 3 = 0$

$- 2 x (3 x + 1) (3 x - 1) + 9 u^{2} - 1 u - 27 u + 3 = 0$

Step 7.1.8.2

Factor out the greatest common factor from each group.

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Step 7.1.8.2.1

Group the first two terms and the last two terms.

$- 2 x (3 x + 1) (3 x - 1) + 9 u^{2} - 1 u - 27 u + 3 = 0$

Step 7.1.8.2.2

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

$- 2 x (3 x + 1) (3 x - 1) + u (9 u - 1) - 3 (9 u - 1) = 0$

$- 2 x (3 x + 1) (3 x - 1) + u (9 u - 1) - 3 (9 u - 1) = 0$

Step 7.1.8.3

Factor the polynomial by factoring out the greatest common factor, $9 u - 1$ .

$- 2 x (3 x + 1) (3 x - 1) + (9 u - 1) (u - 3) = 0$

$- 2 x (3 x + 1) (3 x - 1) + (9 u - 1) (u - 3) = 0$

Step 7.1.9

Replace all occurrences of $u$ with $x^{2}$ .

$- 2 x (3 x + 1) (3 x - 1) + (9 x^{2} - 1) (x^{2} - 3) = 0$

Step 7.1.10

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$- 2 x (3 x + 1) (3 x - 1) + ({(3 x)}^{2} - 1) (x^{2} - 3) = 0$

Step 7.1.11

Rewrite $1$ as $1^{2}$ .

$- 2 x (3 x + 1) (3 x - 1) + ({(3 x)}^{2} - 1^{2}) (x^{2} - 3) = 0$

Step 7.1.12

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 1$ .

$- 2 x (3 x + 1) (3 x - 1) + (3 x + 1) (3 x - 1) (x^{2} - 3) = 0$

Step 7.1.13

Factor $(3 x + 1) (3 x - 1)$ out of $- 2 x (3 x + 1) (3 x - 1) + (3 x + 1) (3 x - 1) (x^{2} - 3)$ .

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Step 7.1.13.1

Factor $(3 x + 1) (3 x - 1)$ out of $- 2 x (3 x + 1) (3 x - 1)$ .

$(3 x + 1) (3 x - 1) (- 2 x) + (3 x + 1) (3 x - 1) (x^{2} - 3) = 0$

Step 7.1.13.2

Factor $(3 x + 1) (3 x - 1)$ out of $(3 x + 1) (3 x - 1) (- 2 x) + (3 x + 1) (3 x - 1) (x^{2} - 3)$ .

$(3 x + 1) (3 x - 1) (- 2 x + x^{2} - 3) = 0$

$(3 x + 1) (3 x - 1) (- 2 x + x^{2} - 3) = 0$

Step 7.1.14

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$(3 x + 1) (3 x - 1) (- 2 u + u^{2} - 3) = 0$

Step 7.1.15

Factor $- 2 u + u^{2} - 3$ using the AC method.

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Step 7.1.15.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 7.1.15.2

Write the factored form using these integers.

$(3 x + 1) (3 x - 1) ((u - 3) (u + 1)) = 0$

$(3 x + 1) (3 x - 1) ((u - 3) (u + 1)) = 0$

Step 7.1.16

Factor.

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Step 7.1.16.1

Replace all occurrences of $u$ with $x$ .

$(3 x + 1) (3 x - 1) ((x - 3) (x + 1)) = 0$

Step 7.1.16.2

Remove unnecessary parentheses.

$(3 x + 1) (3 x - 1) (x - 3) (x + 1) = 0$

$(3 x + 1) (3 x - 1) (x - 3) (x + 1) = 0$

$(3 x + 1) (3 x - 1) (x - 3) (x + 1) = 0$

Step 7.2

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

$3 x + 1 = 0$

$3 x - 1 = 0$

$x - 3 = 0$

$x + 1 = 0$

Step 7.3

Set $3 x + 1$ equal to $0$ and solve for $x$ .

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Step 7.3.1

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Step 7.3.2

Solve $3 x + 1 = 0$ for $x$ .

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Step 7.3.2.1

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Step 7.3.2.2

Divide each term in $3 x = - 1$ by $3$ and simplify.

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Step 7.3.2.2.1

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Step 7.3.2.2.2

Simplify the left side.

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Step 7.3.2.2.2.1

Cancel the common factor of $3$ .

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Step 7.3.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Step 7.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

$x = \frac{- 1}{3}$

$x = \frac{- 1}{3}$

Step 7.3.2.2.3

Simplify the right side.

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Step 7.3.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

$x = - \frac{1}{3}$

$x = - \frac{1}{3}$

$x = - \frac{1}{3}$

$x = - \frac{1}{3}$

Step 7.4

Set $3 x - 1$ equal to $0$ and solve for $x$ .

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Step 7.4.1

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Step 7.4.2

Solve $3 x - 1 = 0$ for $x$ .

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Step 7.4.2.1

Add $1$ to both sides of the equation.

$3 x = 1$

Step 7.4.2.2

Divide each term in $3 x = 1$ by $3$ and simplify.

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Step 7.4.2.2.1

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Step 7.4.2.2.2

Simplify the left side.

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Step 7.4.2.2.2.1

Cancel the common factor of $3$ .

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Step 7.4.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Step 7.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{3}$

$x = \frac{1}{3}$

$x = \frac{1}{3}$

$x = \frac{1}{3}$

$x = \frac{1}{3}$

$x = \frac{1}{3}$

Step 7.5

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

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Step 7.5.1

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

$x - 3 = 0$

Step 7.5.2

Add $3$ to both sides of the equation.

$x = 3$

$x = 3$

Step 7.6

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 7.6.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 7.6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

$x = - 1$

Step 7.7

The final solution is all the values that make $(3 x + 1) (3 x - 1) (x - 3) (x + 1) = 0$ true.

$x = - \frac{1}{3}, \frac{1}{3}, 3, - 1$

$x = - \frac{1}{3}, \frac{1}{3}, 3, - 1$

Step 8

The polynomial can be written as a set of linear factors.

$(x - 1) (x + \frac{1}{3}) (x - \frac{1}{3}) (x - 3) (x + 1)$

Step 9

These are the roots (zeros) of the polynomial $9 x^{5} - 27 x^{4} - 10 x^{3} + 30 x^{2} + x - 3$ .

$x = 1, - \frac{1}{3}, \frac{1}{3}, 3, - 1$

Step 10

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$