Precalculus Examples

$2 x^{4} - x^{3} - 19 x^{2} + 9 x + 9$

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.

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

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 3, \pm \frac{3}{2}, \pm 9, \pm \frac{9}{2}$

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.

$2 {(1)}^{4} - {(1)}^{3} - 19 {(1)}^{2} + 9 (1) + 9$

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

Simplify each term.

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Step 4.1.1

One to any power is one.

$2 \cdot 1 - {(1)}^{3} - 19 {(1)}^{2} + 9 (1) + 9$

Step 4.1.2

Multiply $2$ by $1$ .

$2 - {(1)}^{3} - 19 {(1)}^{2} + 9 (1) + 9$

Step 4.1.3

One to any power is one.

$2 - 1 \cdot 1 - 19 {(1)}^{2} + 9 (1) + 9$

Step 4.1.4

Multiply $- 1$ by $1$ .

$2 - 1 - 19 {(1)}^{2} + 9 (1) + 9$

Step 4.1.5

One to any power is one.

$2 - 1 - 19 \cdot 1 + 9 (1) + 9$

Step 4.1.6

Multiply $- 19$ by $1$ .

$2 - 1 - 19 + 9 (1) + 9$

Step 4.1.7

Multiply $9$ by $1$ .

$2 - 1 - 19 + 9 + 9$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $1$ from $2$ .

$1 - 19 + 9 + 9$

Step 4.2.2

Subtract $19$ from $1$ .

$- 18 + 9 + 9$

Step 4.2.3

Add $- 18$ and $9$ .

$- 9 + 9$

Step 4.2.4

Add $- 9$ and $9$ .

$0$

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.

$\frac{2 x^{4} - x^{3} - 19 x^{2} + 9 x + 9}{x - 1}$

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.

$1$	$2$	$- 1$	$- 19$	$9$	$9$

Step 6.2

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

$1$	$2$	$- 1$	$- 19$	$9$	$9$

	$2$

Step 6.3

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 $(- 1)$ .

$1$	$2$	$- 1$	$- 19$	$9$	$9$
		$2$
	$2$

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.

$1$	$2$	$- 1$	$- 19$	$9$	$9$
		$2$
	$2$	$1$

Step 6.5

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(- 19)$ .

$1$	$2$	$- 1$	$- 19$	$9$	$9$
		$2$	$1$
	$2$	$1$

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.

$1$	$2$	$- 1$	$- 19$	$9$	$9$
		$2$	$1$
	$2$	$1$	$- 18$

Step 6.7

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 $(9)$ .

$1$	$2$	$- 1$	$- 19$	$9$	$9$
		$2$	$1$	$- 18$
	$2$	$1$	$- 18$

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.

$1$	$2$	$- 1$	$- 19$	$9$	$9$
		$2$	$1$	$- 18$
	$2$	$1$	$- 18$	$- 9$

Step 6.9

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 $(9)$ .

$1$	$2$	$- 1$	$- 19$	$9$	$9$
		$2$	$1$	$- 18$	$- 9$
	$2$	$1$	$- 18$	$- 9$

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.

$1$	$2$	$- 1$	$- 19$	$9$	$9$
		$2$	$1$	$- 18$	$- 9$
	$2$	$1$	$- 18$	$- 9$	$0$

Step 6.11

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

$2 x^{3} + 1 x^{2} + (- 18) x - 9$

Step 6.12

Simplify the quotient polynomial.

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

Step 7

Factor out the greatest common factor from each group.

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Step 7.1

Group the first two terms and the last two terms.

$(x - 1) (2 x^{3} + x^{2}) - 18 x - 9$

Step 7.2

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

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

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

Step 8

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

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

Step 9

Rewrite $9$ as $3^{2}$ .

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

Step 10

Factor.

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Step 10.1

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 $b = 3$ .

$(x - 1) + (2 x + 1) ((x + 3) (x - 3))$

Step 10.2

Remove unnecessary parentheses.

$(x - 1) + (2 x + 1) (x + 3) (x - 3)$

$(x - 1) (2 x + 1) (x + 3) (x - 3)$

Step 11

Factor the left side of the equation.

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Step 11.1

Regroup terms.

$- x^{3} + 9 x + 2 x^{4} - 19 x^{2} + 9 = 0$

Step 11.2

Factor $- x$ out of $- x^{3} + 9 x$ .

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Step 11.2.1

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

$- x \cdot x^{2} + 9 x + 2 x^{4} - 19 x^{2} + 9 = 0$

Step 11.2.2

Factor $- x$ out of $9 x$ .

$- x \cdot x^{2} - x \cdot - 9 + 2 x^{4} - 19 x^{2} + 9 = 0$

Step 11.2.3

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

$- x (x^{2} - 9) + 2 x^{4} - 19 x^{2} + 9 = 0$

Step 11.3

Rewrite $9$ as $3^{2}$ .

$- x (x^{2} - 3^{2}) + 2 x^{4} - 19 x^{2} + 9 = 0$

Step 11.4

Factor.

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Step 11.4.1

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 $b = 3$ .

$- x ((x + 3) (x - 3)) + 2 x^{4} - 19 x^{2} + 9 = 0$

Step 11.4.2

Remove unnecessary parentheses.

$- x (x + 3) (x - 3) + 2 x^{4} - 19 x^{2} + 9 = 0$

Step 11.5

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

$- x (x + 3) (x - 3) + 2 {(x^{2})}^{2} - 19 x^{2} + 9 = 0$

Step 11.6

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

$- x (x + 3) (x - 3) + 2 u^{2} - 19 u + 9 = 0$

Step 11.7

Factor by grouping.

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Step 11.7.1

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 $a \cdot c = 2 \cdot 9 = 18$ and whose sum is $b = - 19$ .

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Step 11.7.1.1

Factor $- 19$ out of $- 19 u$ .

$- x (x + 3) (x - 3) + 2 u^{2} - 19 u + 9 = 0$

Step 11.7.1.2

Rewrite $- 19$ as $- 1$ plus $- 18$

$- x (x + 3) (x - 3) + 2 u^{2} + (- 1 - 18) u + 9 = 0$

Step 11.7.1.3

Apply the distributive property.

$- x (x + 3) (x - 3) + 2 u^{2} - 1 u - 18 u + 9 = 0$

Step 11.7.2

Factor out the greatest common factor from each group.

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Step 11.7.2.1

Group the first two terms and the last two terms.

$- x (x + 3) (x - 3) + 2 u^{2} - 1 u - 18 u + 9 = 0$

Step 11.7.2.2

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

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

Step 11.7.3

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

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

Step 11.8

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

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

Step 11.9

Rewrite $9$ as $3^{2}$ .

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

Step 11.10

Factor.

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Step 11.10.1

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 $b = 3$ .

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

Step 11.10.2

Remove unnecessary parentheses.

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

Step 11.11

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

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Step 11.11.1

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

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

Step 11.11.2

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

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

Step 11.11.3

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

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

Step 11.12

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

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

Step 11.13

Factor by grouping.

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Step 11.13.1

Reorder terms.

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

Step 11.13.2

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 $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = - 1$ .

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Step 11.13.2.1

Factor $- 1$ out of $- u$ .

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

Step 11.13.2.2

Rewrite $- 1$ as $1$ plus $- 2$

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

Step 11.13.2.3

Apply the distributive property.

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

Step 11.13.2.4

Multiply $u$ by $1$ .

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

Step 11.13.3

Factor out the greatest common factor from each group.

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Step 11.13.3.1

Group the first two terms and the last two terms.

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

Step 11.13.3.2

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

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

Step 11.13.4

Factor the polynomial by factoring out the greatest common factor, $2 u + 1$ .

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

Step 11.14

Factor.

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Step 11.14.1

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

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

Step 11.14.2

Remove unnecessary parentheses.

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

Step 12

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

$x + 3 = 0$

$x - 3 = 0$

$2 x + 1 = 0$

$x - 1 = 0$

Step 13

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

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Step 13.1

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

$x + 3 = 0$

Step 13.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 14

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

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Step 14.1

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

$x - 3 = 0$

Step 14.2

Add $3$ to both sides of the equation.

$x = 3$

Step 15

Set $2 x + 1$ equal to $0$ and solve for $x$ .

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Step 15.1

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Step 15.2

Solve $2 x + 1 = 0$ for $x$ .

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Step 15.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 15.2.2

Divide each term in $2 x = - 1$ by $2$ and simplify.

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Step 15.2.2.1

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Step 15.2.2.2

Simplify the left side.

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Step 15.2.2.2.1

Cancel the common factor of $2$ .

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Step 15.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Step 15.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Step 15.2.2.3

Simplify the right side.

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Step 15.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 16

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

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Step 16.1

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

$x - 1 = 0$

Step 16.2

Add $1$ to both sides of the equation.

$x = 1$

Step 17

The final solution is all the values that make $(x + 3) (x - 3) (2 x + 1) (x - 1) = 0$ true.

$x = - 3, 3, - \frac{1}{2}, 1$

Step 18