Precalculus Examples

$x^{5} + 3 x^{4} - 9 x^{3} - 31 x^{2} + 36$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36$

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.

${(1)}^{5} + 3 {(1)}^{4} - 9 {(1)}^{3} - 31 {(1)}^{2} + 36$

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.

$1 + 3 {(1)}^{4} - 9 {(1)}^{3} - 31 {(1)}^{2} + 36$

Step 4.1.2

One to any power is one.

$1 + 3 \cdot 1 - 9 {(1)}^{3} - 31 {(1)}^{2} + 36$

Step 4.1.3

Multiply $3$ by $1$ .

$1 + 3 - 9 {(1)}^{3} - 31 {(1)}^{2} + 36$

Step 4.1.4

One to any power is one.

$1 + 3 - 9 \cdot 1 - 31 {(1)}^{2} + 36$

Step 4.1.5

Multiply $- 9$ by $1$ .

$1 + 3 - 9 - 31 {(1)}^{2} + 36$

Step 4.1.6

One to any power is one.

$1 + 3 - 9 - 31 \cdot 1 + 36$

Step 4.1.7

Multiply $- 31$ by $1$ .

$1 + 3 - 9 - 31 + 36$

$1 + 3 - 9 - 31 + 36$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Add $1$ and $3$ .

$4 - 9 - 31 + 36$

Step 4.2.2

Subtract $9$ from $4$ .

$- 5 - 31 + 36$

Step 4.2.3

Subtract $31$ from $- 5$ .

$- 36 + 36$

Step 4.2.4

Add $- 36$ and $36$ .

$0$

$0$

$0$

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.

$\frac{x^{5} + 3 x^{4} - 9 x^{3} - 31 x^{2} + 36}{x - 1}$

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.

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$

Step 6.2

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

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$

	$1$

Step 6.3

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 $(3)$ .

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$
		$1$
	$1$

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.

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$
		$1$
	$1$	$4$

Step 6.5

Multiply the newest entry in the result $(4)$ by the divisor $(1)$ and place the result of $(4)$ under the next term in the dividend $(- 9)$ .

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$
		$1$	$4$
	$1$	$4$

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.

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$
		$1$	$4$
	$1$	$4$	$- 5$

Step 6.7

Multiply the newest entry in the result $(- 5)$ by the divisor $(1)$ and place the result of $(- 5)$ under the next term in the dividend $(- 31)$ .

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$
		$1$	$4$	$- 5$
	$1$	$4$	$- 5$

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.

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$
		$1$	$4$	$- 5$
	$1$	$4$	$- 5$	$- 36$

Step 6.9

Multiply the newest entry in the result $(- 36)$ by the divisor $(1)$ and place the result of $(- 36)$ under the next term in the dividend $(0)$ .

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$
		$1$	$4$	$- 5$	$- 36$
	$1$	$4$	$- 5$	$- 36$

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.

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$
		$1$	$4$	$- 5$	$- 36$
	$1$	$4$	$- 5$	$- 36$	$- 36$

Step 6.11

Multiply the newest entry in the result $(- 36)$ by the divisor $(1)$ and place the result of $(- 36)$ under the next term in the dividend $(36)$ .

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$
		$1$	$4$	$- 5$	$- 36$	$- 36$
	$1$	$4$	$- 5$	$- 36$	$- 36$

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.

$1$	$1$	$3$	$- 9$	$- 31$	$0$	$36$
		$1$	$4$	$- 5$	$- 36$	$- 36$
	$1$	$4$	$- 5$	$- 36$	$- 36$	$0$

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.

$1 x^{4} + 4 x^{3} - 5 x^{2} + - 36 x - 36$

Step 6.14

Simplify the quotient polynomial.

$x^{4} + 4 x^{3} - 5 x^{2} - 36 x - 36$

$x^{4} + 4 x^{3} - 5 x^{2} - 36 x - 36$

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.

$4 x^{3} - 36 x + x^{4} - 5 x^{2} - 36 = 0$

Step 7.1.2

Factor $4 x$ out of $4 x^{3} - 36 x$ .

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Step 7.1.2.1

Factor $4 x$ out of $4 x^{3}$ .

$4 x (x^{2}) - 36 x + x^{4} - 5 x^{2} - 36 = 0$

Step 7.1.2.2

Factor $4 x$ out of $- 36 x$ .

$4 x (x^{2}) + 4 x (- 9) + x^{4} - 5 x^{2} - 36 = 0$

Step 7.1.2.3

Factor $4 x$ out of $4 x (x^{2}) + 4 x (- 9)$ .

$4 x (x^{2} - 9) + x^{4} - 5 x^{2} - 36 = 0$

$4 x (x^{2} - 9) + x^{4} - 5 x^{2} - 36 = 0$

Step 7.1.3

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

$4 x (x^{2} - 3^{2}) + x^{4} - 5 x^{2} - 36 = 0$

Step 7.1.4

Factor.

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Step 7.1.4.1

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 $b = 3$ .

$4 x ((x + 3) (x - 3)) + x^{4} - 5 x^{2} - 36 = 0$

Step 7.1.4.2

Remove unnecessary parentheses.

$4 x (x + 3) (x - 3) + x^{4} - 5 x^{2} - 36 = 0$

$4 x (x + 3) (x - 3) + x^{4} - 5 x^{2} - 36 = 0$

Step 7.1.5

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

$4 x (x + 3) (x - 3) + {(x^{2})}^{2} - 5 x^{2} - 36 = 0$

Step 7.1.6

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

$4 x (x + 3) (x - 3) + u^{2} - 5 u - 36 = 0$

Step 7.1.7

Factor $u^{2} - 5 u - 36$ using the AC method.

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Step 7.1.7.1

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 $- 36$ and whose sum is $- 5$ .

$- 9, 4$

Step 7.1.7.2

Write the factored form using these integers.

$4 x (x + 3) (x - 3) + (u - 9) (u + 4) = 0$

$4 x (x + 3) (x - 3) + (u - 9) (u + 4) = 0$

Step 7.1.8

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

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

Step 7.1.9

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

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

Step 7.1.10

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 $b = 3$ .

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

Step 7.1.11

Factor $(x + 3) (x - 3)$ out of $4 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} + 4)$ .

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Step 7.1.11.1

Factor $(x + 3) (x - 3)$ out of $4 x (x + 3) (x - 3)$ .

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

Step 7.1.11.2

Factor $(x + 3) (x - 3)$ out of $(x + 3) (x - 3) (4 x) + (x + 3) (x - 3) (x^{2} + 4)$ .

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

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

Step 7.1.12

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

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

Step 7.1.13

Factor using the perfect square rule.

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Step 7.1.13.1

Rearrange terms.

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

Step 7.1.13.2

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

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

Step 7.1.13.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 u = 2 \cdot u \cdot 2$

Step 7.1.13.4

Rewrite the polynomial.

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

Step 7.1.13.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 2$ .

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

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

Step 7.1.14

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

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

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

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 $0$ .

$x + 3 = 0$

$x - 3 = 0$

${(x + 2)}^{2} = 0$

Step 7.3

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

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Step 7.3.1

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

$x + 3 = 0$

Step 7.3.2

Subtract $3$ from both sides of the equation.

$x = - 3$

$x = - 3$

Step 7.4

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

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Step 7.4.1

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

$x - 3 = 0$

Step 7.4.2

Add $3$ to both sides of the equation.

$x = 3$

$x = 3$

Step 7.5

Set ${(x + 2)}^{2}$ equal to $0$ and solve for $x$ .

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Step 7.5.1

Set ${(x + 2)}^{2}$ equal to $0$ .

${(x + 2)}^{2} = 0$

Step 7.5.2

Solve ${(x + 2)}^{2} = 0$ for $x$ .

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Step 7.5.2.1

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

$x + 2 = 0$

Step 7.5.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

$x = - 2$

$x = - 2$

Step 7.6

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

$x = - 3, 3, - 2$

$x = - 3, 3, - 2$

Step 8

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

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

Step 9

These are the roots (zeros) of the polynomial $x^{5} + 3 x^{4} - 9 x^{3} - 31 x^{2} + 36$ .

$x = 1, - 3, 3, - 2$

Step 10

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