Precalculus Examples

$2 x^{4} - x^{3} + 49 x^{2} - 25 x - 25$

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.

$p = \pm 1, \pm 5, \pm 25$

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 5, \pm \frac{5}{2}, \pm 25, \pm \frac{25}{2}$

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.

$2 {(1)}^{4} - {(1)}^{3} + 49 {(1)}^{2} - 25 \cdot 1 - 25$

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.

$2 \cdot 1 - {(1)}^{3} + 49 {(1)}^{2} - 25 \cdot 1 - 25$

Step 4.1.2

Multiply $2$ by $1$ .

$2 - {(1)}^{3} + 49 {(1)}^{2} - 25 \cdot 1 - 25$

Step 4.1.3

One to any power is one.

$2 - 1 \cdot 1 + 49 {(1)}^{2} - 25 \cdot 1 - 25$

Step 4.1.4

Multiply $- 1$ by $1$ .

$2 - 1 + 49 {(1)}^{2} - 25 \cdot 1 - 25$

Step 4.1.5

One to any power is one.

$2 - 1 + 49 \cdot 1 - 25 \cdot 1 - 25$

Step 4.1.6

Multiply $49$ by $1$ .

$2 - 1 + 49 - 25 \cdot 1 - 25$

Step 4.1.7

Multiply $- 25$ by $1$ .

$2 - 1 + 49 - 25 - 25$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $1$ from $2$ .

$1 + 49 - 25 - 25$

Step 4.2.2

Add $1$ and $49$ .

$50 - 25 - 25$

Step 4.2.3

Subtract $25$ from $50$ .

$25 - 25$

Step 4.2.4

Subtract $25$ from $25$ .

$0$

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.

$\frac{2 x^{4} - x^{3} + 49 x^{2} - 25 x - 25}{x - 1}$

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.

$1$	$2$	$- 1$	$49$	$- 25$	$- 25$

Step 6.2

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

$1$	$2$	$- 1$	$49$	$- 25$	$- 25$

	$2$

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 $(- 1)$ .

$1$	$2$	$- 1$	$49$	$- 25$	$- 25$
		$2$
	$2$

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.

$1$	$2$	$- 1$	$49$	$- 25$	$- 25$
		$2$
	$2$	$1$

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 $(49)$ .

$1$	$2$	$- 1$	$49$	$- 25$	$- 25$
		$2$	$1$
	$2$	$1$

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.

$1$	$2$	$- 1$	$49$	$- 25$	$- 25$
		$2$	$1$
	$2$	$1$	$50$

Step 6.7

Multiply the newest entry in the result $(50)$ by the divisor $(1)$ and place the result of $(50)$ under the next term in the dividend $(- 25)$ .

$1$	$2$	$- 1$	$49$	$- 25$	$- 25$
		$2$	$1$	$50$
	$2$	$1$	$50$

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.

$1$	$2$	$- 1$	$49$	$- 25$	$- 25$
		$2$	$1$	$50$
	$2$	$1$	$50$	$25$

Step 6.9

Multiply the newest entry in the result $(25)$ by the divisor $(1)$ and place the result of $(25)$ under the next term in the dividend $(- 25)$ .

$1$	$2$	$- 1$	$49$	$- 25$	$- 25$
		$2$	$1$	$50$	$25$
	$2$	$1$	$50$	$25$

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.

$1$	$2$	$- 1$	$49$	$- 25$	$- 25$
		$2$	$1$	$50$	$25$
	$2$	$1$	$50$	$25$	$0$

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.

$2 x^{3} + 1 x^{2} + (50) x + 25$

Step 6.12

Simplify the quotient polynomial.

$2 x^{3} + x^{2} + 50 x + 25$

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.

$(x - 1) (2 x^{3} + x^{2}) + 50 x + 25$

Step 7.2

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

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

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

Step 8

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

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

Step 9

Factor the left side of the equation.

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Step 9.1

Regroup terms.

$- x^{3} - 25 x + 2 x^{4} + 49 x^{2} - 25 = 0$

Step 9.2

Factor $- x$ out of $- x^{3} - 25 x$ .

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Step 9.2.1

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

$- x \cdot x^{2} - 25 x + 2 x^{4} + 49 x^{2} - 25 = 0$

Step 9.2.2

Factor $- x$ out of $- 25 x$ .

$- x \cdot x^{2} - x \cdot 25 + 2 x^{4} + 49 x^{2} - 25 = 0$

Step 9.2.3

Factor $- x$ out of $- x (x^{2}) - x (25)$ .

$- x (x^{2} + 25) + 2 x^{4} + 49 x^{2} - 25 = 0$

Step 9.3

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

$- x (x^{2} + 25) + 2 {(x^{2})}^{2} + 49 x^{2} - 25 = 0$

Step 9.4

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

$- x (x^{2} + 25) + 2 u^{2} + 49 u - 25 = 0$

Step 9.5

Factor by grouping.

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Step 9.5.1

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 $a \cdot c = 2 \cdot - 25 = - 50$ and whose sum is $b = 49$ .

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Step 9.5.1.1

Factor $49$ out of $49 u$ .

$- x (x^{2} + 25) + 2 u^{2} + 49 (u) - 25 = 0$

Step 9.5.1.2

Rewrite $49$ as $- 1$ plus $50$

$- x (x^{2} + 25) + 2 u^{2} + (- 1 + 50) u - 25 = 0$

Step 9.5.1.3

Apply the distributive property.

$- x (x^{2} + 25) + 2 u^{2} - 1 u + 50 u - 25 = 0$

Step 9.5.2

Factor out the greatest common factor from each group.

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Step 9.5.2.1

Group the first two terms and the last two terms.

$- x (x^{2} + 25) + 2 u^{2} - 1 u + 50 u - 25 = 0$

Step 9.5.2.2

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

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

Step 9.5.3

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

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

Step 9.6

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

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

Step 9.7

Factor $x^{2} + 25$ out of $- x (x^{2} + 25) + (2 x^{2} - 1) (x^{2} + 25)$ .

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Step 9.7.1

Factor $x^{2} + 25$ out of $- x (x^{2} + 25)$ .

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

Step 9.7.2

Factor $x^{2} + 25$ out of $(2 x^{2} - 1) (x^{2} + 25)$ .

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

Step 9.7.3

Factor $x^{2} + 25$ out of $(x^{2} + 25) (- x) + (x^{2} + 25) (2 x^{2} - 1)$ .

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

Step 9.8

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

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

Step 9.9

Factor by grouping.

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Step 9.9.1

Reorder terms.

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

Step 9.9.2

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 9.9.2.1

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

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

Step 9.9.2.2

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

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

Step 9.9.2.3

Apply the distributive property.

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

Step 9.9.2.4

Multiply $u$ by $1$ .

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

Step 9.9.3

Factor out the greatest common factor from each group.

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Step 9.9.3.1

Group the first two terms and the last two terms.

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

Step 9.9.3.2

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

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

Step 9.9.4

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

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

Step 9.10

Factor.

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Step 9.10.1

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

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

Step 9.10.2

Remove unnecessary parentheses.

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

Step 10

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

$x^{2} + 25 = 0$

$2 x + 1 = 0$

$x - 1 = 0$

Step 11

Set $x^{2} + 25$ equal to $0$ and solve for $x$ .

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Step 11.1

Set $x^{2} + 25$ equal to $0$ .

$x^{2} + 25 = 0$

Step 11.2

Solve $x^{2} + 25 = 0$ for $x$ .

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Step 11.2.1

Subtract $25$ from both sides of the equation.

$x^{2} = - 25$

Step 11.2.2

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

$x = \pm \sqrt{- 25}$

Step 11.2.3

Simplify $\pm \sqrt{- 25}$ .

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Step 11.2.3.1

Rewrite $- 25$ as $- 1 (25)$ .

$x = \pm \sqrt{- 1 (25)}$

Step 11.2.3.2

Rewrite $\sqrt{- 1 (25)}$ as $\sqrt{- 1} \cdot \sqrt{25}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{25}$

Step 11.2.3.3

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

$x = \pm i \cdot \sqrt{25}$

Step 11.2.3.4

Rewrite $25$ as $5^{2}$ .

$x = \pm i \cdot \sqrt{5^{2}}$

Step 11.2.3.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 5$

Step 11.2.3.6

Move $5$ to the left of $i$ .

$x = \pm 5 i$

Step 11.2.4

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

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Step 11.2.4.1

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

$x = 5 i$

Step 11.2.4.2

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

$x = - 5 i$

Step 11.2.4.3

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

$x = 5 i, - 5 i$

Step 12

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

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Step 12.1

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

$2 x + 1 = 0$

Step 12.2

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

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Step 12.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 12.2.2

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

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Step 12.2.2.1

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

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

Step 12.2.2.2

Simplify the left side.

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Step 12.2.2.2.1

Cancel the common factor of $2$ .

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Step 12.2.2.2.1.1

Cancel the common factor.

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

Step 12.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 12.2.2.3

Simplify the right side.

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Step 12.2.2.3.1

Move the negative in front of the fraction.

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

Step 13

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

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Step 13.1

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

$x - 1 = 0$

Step 13.2

Add $1$ to both sides of the equation.

$x = 1$

Step 14

The final solution is all the values that make $(x^{2} + 25) (2 x + 1) (x - 1) = 0$ true.

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

Step 15