Precalculus Examples

$4 x^{5} - 8 x^{4} - 5 x^{3} + 10 x^{2} + x - 2$

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.

$p = \pm 1, \pm 2$

$q = \pm 1, \pm 2, \pm 4$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 2$

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.

$4 {(1)}^{5} - 8 {(1)}^{4} - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

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.

$4 {(1)}^{5} - 8 {(1)}^{4} - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

Step 4.2

Simplify each term.

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Step 4.2.1

One to any power is one.

$4 \cdot 1 - 8 {(1)}^{4} - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

Step 4.2.2

Multiply $4$ by $1$ .

$4 - 8 {(1)}^{4} - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

Step 4.2.3

One to any power is one.

$4 - 8 \cdot 1 - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

Step 4.2.4

Multiply $- 8$ by $1$ .

$4 - 8 - 5 {(1)}^{3} + 10 {(1)}^{2} + 1 - 2$

Step 4.2.5

One to any power is one.

$4 - 8 - 5 \cdot 1 + 10 {(1)}^{2} + 1 - 2$

Step 4.2.6

Multiply $- 5$ by $1$ .

$4 - 8 - 5 + 10 {(1)}^{2} + 1 - 2$

Step 4.2.7

One to any power is one.

$4 - 8 - 5 + 10 \cdot 1 + 1 - 2$

Step 4.2.8

Multiply $10$ by $1$ .

$4 - 8 - 5 + 10 + 1 - 2$

$4 - 8 - 5 + 10 + 1 - 2$

Step 4.3

Simplify by adding and subtracting.

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Step 4.3.1

Subtract $8$ from $4$ .

$- 4 - 5 + 10 + 1 - 2$

Step 4.3.2

Subtract $5$ from $- 4$ .

$- 9 + 10 + 1 - 2$

Step 4.3.3

Add $- 9$ and $10$ .

$1 + 1 - 2$

Step 4.3.4

Add $1$ and $1$ .

$2 - 2$

Step 4.3.5

Subtract $2$ from $2$ .

$0$

$0$

$0$

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.

$\frac{4 x^{5} - 8 x^{4} - 5 x^{3} + 10 x^{2} + x - 2}{x - 1}$

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.

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$

Step 6.2

The first number in the dividend $(4)$ is put into the first position of the result area (below the horizontal line).

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$

	$4$

Step 6.3

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 $(- 8)$ .

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
		$4$
	$4$

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.

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
		$4$
	$4$	$- 4$

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 $(- 5)$ .

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
		$4$	$- 4$
	$4$	$- 4$

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.

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
		$4$	$- 4$
	$4$	$- 4$	$- 9$

Step 6.7

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 $(10)$ .

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
		$4$	$- 4$	$- 9$
	$4$	$- 4$	$- 9$

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.

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
		$4$	$- 4$	$- 9$
	$4$	$- 4$	$- 9$	$1$

Step 6.9

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 $(1)$ .

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
		$4$	$- 4$	$- 9$	$1$
	$4$	$- 4$	$- 9$	$1$

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.

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
		$4$	$- 4$	$- 9$	$1$
	$4$	$- 4$	$- 9$	$1$	$2$

Step 6.11

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 $(- 2)$ .

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
		$4$	$- 4$	$- 9$	$1$	$2$
	$4$	$- 4$	$- 9$	$1$	$2$

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.

$1$	$4$	$- 8$	$- 5$	$10$	$1$	$- 2$
		$4$	$- 4$	$- 9$	$1$	$2$
	$4$	$- 4$	$- 9$	$1$	$2$	$0$

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.

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

Step 6.14

Simplify the quotient polynomial.

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

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

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.

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

Step 7.1.2

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

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Step 7.1.2.1

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

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

Step 7.1.2.2

Rewrite $x$ as $1 x$ .

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

Step 7.1.2.3

Factor $- x$ out of $1 x$ .

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

Step 7.1.2.4

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

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

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

Step 7.1.3

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

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

Step 7.1.4

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

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

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 $a = 2 x$ and $b = 1$ .

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

Step 7.1.5.2

Remove unnecessary parentheses.

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

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

Step 7.1.6

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

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

Step 7.1.7

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

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

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 $a \cdot c = 4 \cdot 2 = 8$ and whose sum is $b = - 9$ .

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Step 7.1.8.1.1

Factor $- 9$ out of $- 9 u$ .

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

Step 7.1.8.1.2

Rewrite $- 9$ as $- 1$ plus $- 8$

$- x (2 x + 1) (2 x - 1) + 4 u^{2} + (- 1 - 8) u + 2 = 0$

Step 7.1.8.1.3

Apply the distributive property.

$- x (2 x + 1) (2 x - 1) + 4 u^{2} - 1 u - 8 u + 2 = 0$

$- x (2 x + 1) (2 x - 1) + 4 u^{2} - 1 u - 8 u + 2 = 0$

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.

$- x (2 x + 1) (2 x - 1) + 4 u^{2} - 1 u - 8 u + 2 = 0$

Step 7.1.8.2.2

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

$- x (2 x + 1) (2 x - 1) + u (4 u - 1) - 2 (4 u - 1) = 0$

$- x (2 x + 1) (2 x - 1) + u (4 u - 1) - 2 (4 u - 1) = 0$

Step 7.1.8.3

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

$- x (2 x + 1) (2 x - 1) + (4 u - 1) (u - 2) = 0$

$- x (2 x + 1) (2 x - 1) + (4 u - 1) (u - 2) = 0$

Step 7.1.9

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

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

Step 7.1.10

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

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

Step 7.1.11

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

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

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 $a = 2 x$ and $b = 1$ .

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

Step 7.1.13

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

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Step 7.1.13.1

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

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

Step 7.1.13.2

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

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

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

Step 7.1.14

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

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

Step 7.1.15

Factor $- u + u^{2} - 2$ 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 $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Step 7.1.15.2

Write the factored form using these integers.

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

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

Step 7.1.16

Factor.

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Step 7.1.16.1

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

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

Step 7.1.16.2

Remove unnecessary parentheses.

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

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

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

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 $0$ .

$2 x + 1 = 0$

$2 x - 1 = 0$

$x - 2 = 0$

$x + 1 = 0$

Step 7.3

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

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Step 7.3.1

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

$2 x + 1 = 0$

Step 7.3.2

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

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Step 7.3.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 7.3.2.2

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

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Step 7.3.2.2.1

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

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

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 $2$ .

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Step 7.3.2.2.2.1.1

Cancel the common factor.

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

Step 7.3.2.2.2.1.2

Divide $x$ by $1$ .

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

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

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

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.

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

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

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

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

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

Step 7.4

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

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Step 7.4.1

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

$2 x - 1 = 0$

Step 7.4.2

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

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Step 7.4.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 7.4.2.2

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

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Step 7.4.2.2.1

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

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

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 $2$ .

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Step 7.4.2.2.2.1.1

Cancel the common factor.

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

Step 7.4.2.2.2.1.2

Divide $x$ by $1$ .

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

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

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

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

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

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

Step 7.5

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

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Step 7.5.1

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

$x - 2 = 0$

Step 7.5.2

Add $2$ to both sides of the equation.

$x = 2$

$x = 2$

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 $(2 x + 1) (2 x - 1) (x - 2) (x + 1) = 0$ true.

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

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

Step 8

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

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

Step 9

These are the roots (zeros) of the polynomial $4 x^{5} - 8 x^{4} - 5 x^{3} + 10 x^{2} + x - 2$ .

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

Step 10

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