Precalculus Examples

$5 x^{5} - 36 x^{4} + 62 x^{3} - 46 x^{2} + 57 x - 10$

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.

$p = \pm 1, \pm 2, \pm 5, \pm 10$

$q = \pm 1, \pm 5$

Step 2

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

$\pm 1, \pm \frac{1}{5}, \pm 2, \pm \frac{2}{5}, \pm 5, \pm 10$

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.

$5 {(2)}^{5} - 36 {(2)}^{4} + 62 {(2)}^{3} - 46 {(2)}^{2} + 57 (2) - 10$

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

Raise $2$ to the power of $5$ .

$5 \cdot 32 - 36 {(2)}^{4} + 62 {(2)}^{3} - 46 {(2)}^{2} + 57 (2) - 10$

Step 4.1.2

Multiply $5$ by $32$ .

$160 - 36 {(2)}^{4} + 62 {(2)}^{3} - 46 {(2)}^{2} + 57 (2) - 10$

Step 4.1.3

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

$160 - 36 \cdot 16 + 62 {(2)}^{3} - 46 {(2)}^{2} + 57 (2) - 10$

Step 4.1.4

Multiply $- 36$ by $16$ .

$160 - 576 + 62 {(2)}^{3} - 46 {(2)}^{2} + 57 (2) - 10$

Step 4.1.5

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

$160 - 576 + 62 \cdot 8 - 46 {(2)}^{2} + 57 (2) - 10$

Step 4.1.6

Multiply $62$ by $8$ .

$160 - 576 + 496 - 46 {(2)}^{2} + 57 (2) - 10$

Step 4.1.7

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

$160 - 576 + 496 - 46 \cdot 4 + 57 (2) - 10$

Step 4.1.8

Multiply $- 46$ by $4$ .

$160 - 576 + 496 - 184 + 57 (2) - 10$

Step 4.1.9

Multiply $57$ by $2$ .

$160 - 576 + 496 - 184 + 114 - 10$

$160 - 576 + 496 - 184 + 114 - 10$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $576$ from $160$ .

$- 416 + 496 - 184 + 114 - 10$

Step 4.2.2

Add $- 416$ and $496$ .

$80 - 184 + 114 - 10$

Step 4.2.3

Subtract $184$ from $80$ .

$- 104 + 114 - 10$

Step 4.2.4

Add $- 104$ and $114$ .

$10 - 10$

Step 4.2.5

Subtract $10$ from $10$ .

$0$

$0$

$0$

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.

$\frac{5 x^{5} - 36 x^{4} + 62 x^{3} - 46 x^{2} + 57 x - 10}{x - 2}$

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.

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$

Step 6.2

The first number in the dividend $(5)$ is put into the first position of the result area (below the horizontal line).

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$

	$5$

Step 6.3

Multiply the newest entry in the result $(5)$ by the divisor $(2)$ and place the result of $(10)$ under the next term in the dividend $(- 36)$ .

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$
		$10$
	$5$

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.

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$
		$10$
	$5$	$- 26$

Step 6.5

Multiply the newest entry in the result $(- 26)$ by the divisor $(2)$ and place the result of $(- 52)$ under the next term in the dividend $(62)$ .

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$
		$10$	$- 52$
	$5$	$- 26$

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.

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$
		$10$	$- 52$
	$5$	$- 26$	$10$

Step 6.7

Multiply the newest entry in the result $(10)$ by the divisor $(2)$ and place the result of $(20)$ under the next term in the dividend $(- 46)$ .

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$
		$10$	$- 52$	$20$
	$5$	$- 26$	$10$

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.

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$
		$10$	$- 52$	$20$
	$5$	$- 26$	$10$	$- 26$

Step 6.9

Multiply the newest entry in the result $(- 26)$ by the divisor $(2)$ and place the result of $(- 52)$ under the next term in the dividend $(57)$ .

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$
		$10$	$- 52$	$20$	$- 52$
	$5$	$- 26$	$10$	$- 26$

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.

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$
		$10$	$- 52$	$20$	$- 52$
	$5$	$- 26$	$10$	$- 26$	$5$

Step 6.11

Multiply the newest entry in the result $(5)$ by the divisor $(2)$ and place the result of $(10)$ under the next term in the dividend $(- 10)$ .

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$
		$10$	$- 52$	$20$	$- 52$	$10$
	$5$	$- 26$	$10$	$- 26$	$5$

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.

$2$	$5$	$- 36$	$62$	$- 46$	$57$	$- 10$
		$10$	$- 52$	$20$	$- 52$	$10$
	$5$	$- 26$	$10$	$- 26$	$5$	$0$

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.

$5 x^{4} + - 26 x^{3} + 10 x^{2} + - 26 x + 5$

Step 6.14

Simplify the quotient polynomial.

$5 x^{4} - 26 x^{3} + 10 x^{2} - 26 x + 5$

$5 x^{4} - 26 x^{3} + 10 x^{2} - 26 x + 5$

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.

$- 26 x^{3} - 26 x + 5 x^{4} + 10 x^{2} + 5 = 0$

Step 7.1.2

Factor $- 26 x$ out of $- 26 x^{3} - 26 x$ .

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Step 7.1.2.1

Factor $- 26 x$ out of $- 26 x^{3}$ .

$- 26 x \cdot x^{2} - 26 x + 5 x^{4} + 10 x^{2} + 5 = 0$

Step 7.1.2.2

Factor $- 26 x$ out of $- 26 x$ .

$- 26 x \cdot x^{2} - 26 x \cdot 1 + 5 x^{4} + 10 x^{2} + 5 = 0$

Step 7.1.2.3

Factor $- 26 x$ out of $- 26 x (x^{2}) - 26 x (1)$ .

$- 26 x (x^{2} + 1) + 5 x^{4} + 10 x^{2} + 5 = 0$

$- 26 x (x^{2} + 1) + 5 x^{4} + 10 x^{2} + 5 = 0$

Step 7.1.3

Factor $5$ out of $5 x^{4} + 10 x^{2} + 5$ .

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Step 7.1.3.1

Factor $5$ out of $5 x^{4}$ .

$- 26 x (x^{2} + 1) + 5 (x^{4}) + 10 x^{2} + 5 = 0$

Step 7.1.3.2

Factor $5$ out of $10 x^{2}$ .

$- 26 x (x^{2} + 1) + 5 (x^{4}) + 5 (2 x^{2}) + 5 = 0$

Step 7.1.3.3

Factor $5$ out of $5$ .

$- 26 x (x^{2} + 1) + 5 (x^{4}) + 5 (2 x^{2}) + 5 (1) = 0$

Step 7.1.3.4

Factor $5$ out of $5 (x^{4}) + 5 (2 x^{2})$ .

$- 26 x (x^{2} + 1) + 5 (x^{4} + 2 x^{2}) + 5 (1) = 0$

Step 7.1.3.5

Factor $5$ out of $5 (x^{4} + 2 x^{2}) + 5 (1)$ .

$- 26 x (x^{2} + 1) + 5 (x^{4} + 2 x^{2} + 1) = 0$

$- 26 x (x^{2} + 1) + 5 (x^{4} + 2 x^{2} + 1) = 0$

Step 7.1.4

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

$- 26 x (x^{2} + 1) + 5 ({(x^{2})}^{2} + 2 x^{2} + 1) = 0$

Step 7.1.5

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

$- 26 x (x^{2} + 1) + 5 (u^{2} + 2 u + 1) = 0$

Step 7.1.6

Factor using the perfect square rule.

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Step 7.1.6.1

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

$- 26 x (x^{2} + 1) + 5 (u^{2} + 2 u + 1^{2}) = 0$

Step 7.1.6.2

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

$2 u = 2 \cdot u \cdot 1$

Step 7.1.6.3

Rewrite the polynomial.

$- 26 x (x^{2} + 1) + 5 (u^{2} + 2 \cdot u \cdot 1 + 1^{2}) = 0$

Step 7.1.6.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 1$ .

$- 26 x (x^{2} + 1) + 5 {(u + 1)}^{2} = 0$

$- 26 x (x^{2} + 1) + 5 {(u + 1)}^{2} = 0$

Step 7.1.7

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

$- 26 x (x^{2} + 1) + 5 {(x^{2} + 1)}^{2} = 0$

Step 7.1.8

Factor $x^{2} + 1$ out of $- 26 x (x^{2} + 1) + 5 {(x^{2} + 1)}^{2}$ .

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Step 7.1.8.1

Factor $x^{2} + 1$ out of $- 26 x (x^{2} + 1)$ .

$(x^{2} + 1) (- 26 x) + 5 {(x^{2} + 1)}^{2} = 0$

Step 7.1.8.2

Factor $x^{2} + 1$ out of $5 {(x^{2} + 1)}^{2}$ .

$(x^{2} + 1) (- 26 x) + (x^{2} + 1) (5 (x^{2} + 1)) = 0$

Step 7.1.8.3

Factor $x^{2} + 1$ out of $(x^{2} + 1) (- 26 x) + (x^{2} + 1) (5 (x^{2} + 1))$ .

$(x^{2} + 1) (- 26 x + 5 (x^{2} + 1)) = 0$

$(x^{2} + 1) (- 26 x + 5 (x^{2} + 1)) = 0$

Step 7.1.9

Apply the distributive property.

$(x^{2} + 1) (- 26 x + 5 x^{2} + 5 \cdot 1) = 0$

Step 7.1.10

Multiply $5$ by $1$ .

$(x^{2} + 1) (- 26 x + 5 x^{2} + 5) = 0$

Step 7.1.11

Reorder terms.

$(x^{2} + 1) (5 x^{2} - 26 x + 5) = 0$

Step 7.1.12

Factor.

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Step 7.1.12.1

Factor by grouping.

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Step 7.1.12.1.1

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 $a \cdot c = 5 \cdot 5 = 25$ and whose sum is $b = - 26$ .

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Step 7.1.12.1.1.1

Factor $- 26$ out of $- 26 x$ .

$(x^{2} + 1) (5 x^{2} - 26 x + 5) = 0$

Step 7.1.12.1.1.2

Rewrite $- 26$ as $- 1$ plus $- 25$

$(x^{2} + 1) (5 x^{2} + (- 1 - 25) x + 5) = 0$

Step 7.1.12.1.1.3

Apply the distributive property.

$(x^{2} + 1) (5 x^{2} - 1 x - 25 x + 5) = 0$

$(x^{2} + 1) (5 x^{2} - 1 x - 25 x + 5) = 0$

Step 7.1.12.1.2

Factor out the greatest common factor from each group.

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Step 7.1.12.1.2.1

Group the first two terms and the last two terms.

$(x^{2} + 1) ((5 x^{2} - 1 x) - 25 x + 5) = 0$

Step 7.1.12.1.2.2

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

$(x^{2} + 1) (x (5 x - 1) - 5 (5 x - 1)) = 0$

$(x^{2} + 1) (x (5 x - 1) - 5 (5 x - 1)) = 0$

Step 7.1.12.1.3

Factor the polynomial by factoring out the greatest common factor, $5 x - 1$ .

$(x^{2} + 1) ((5 x - 1) (x - 5)) = 0$

$(x^{2} + 1) ((5 x - 1) (x - 5)) = 0$

Step 7.1.12.2

Remove unnecessary parentheses.

$(x^{2} + 1) (5 x - 1) (x - 5) = 0$

$(x^{2} + 1) (5 x - 1) (x - 5) = 0$

$(x^{2} + 1) (5 x - 1) (x - 5) = 0$

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 $0$ .

$x^{2} + 1 = 0$

$5 x - 1 = 0$

$x - 5 = 0$

Step 7.3

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

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Step 7.3.1

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Step 7.3.2

Solve $x^{2} + 1 = 0$ for $x$ .

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Step 7.3.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 7.3.2.2

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

$x = \pm \sqrt{- 1}$

Step 7.3.2.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 7.3.2.4

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

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Step 7.3.2.4.1

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

$x = i$

Step 7.3.2.4.2

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

$x = - i$

Step 7.3.2.4.3

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

$x = i, - i$

$x = i, - i$

$x = i, - i$

$x = i, - i$

Step 7.4

Set $5 x - 1$ equal to $0$ and solve for $x$ .

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Step 7.4.1

Set $5 x - 1$ equal to $0$ .

$5 x - 1 = 0$

Step 7.4.2

Solve $5 x - 1 = 0$ for $x$ .

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Step 7.4.2.1

Add $1$ to both sides of the equation.

$5 x = 1$

Step 7.4.2.2

Divide each term in $5 x = 1$ by $5$ and simplify.

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Step 7.4.2.2.1

Divide each term in $5 x = 1$ by $5$ .

$\frac{5 x}{5} = \frac{1}{5}$

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 $5$ .

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Step 7.4.2.2.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{1}{5}$

Step 7.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{5}$

$x = \frac{1}{5}$

$x = \frac{1}{5}$

$x = \frac{1}{5}$

$x = \frac{1}{5}$

$x = \frac{1}{5}$

Step 7.5

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 7.5.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 7.5.2

Add $5$ to both sides of the equation.

$x = 5$

$x = 5$

Step 7.6

The final solution is all the values that make $(x^{2} + 1) (5 x - 1) (x - 5) = 0$ true.

$x = i, - i, \frac{1}{5}, 5$

$x = i, - i, \frac{1}{5}, 5$

Step 8

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

$(x - 2) (x - i) (x + i) (x - \frac{1}{5}) (x - 5)$

Step 9

These are the roots (zeros) of the polynomial $5 x^{5} - 36 x^{4} + 62 x^{3} - 46 x^{2} + 57 x - 10$ .

$x = 2, i, - i, \frac{1}{5}, 5$

Step 10

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