Precalculus Examples

$f (x) = 2 x^{4} - x^{3} + 49 x^{2} - 25 x - 25$

Step 1

Set $2 x^{4} - x^{3} + 49 x^{2} - 25 x - 25$ equal to $0$ .

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

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Regroup terms.

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

Step 2.1.2

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

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Step 2.1.2.1

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

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

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

Step 2.1.3

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

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

Step 2.1.4

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

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

Step 2.1.5

Factor by grouping.

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Step 2.1.5.1

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 2.1.5.1.1

Factor $49$ out of $49 u$ .

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

Step 2.1.5.1.2

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

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

Step 2.1.5.1.3

Apply the distributive property.

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

Step 2.1.5.2

Factor out the greatest common factor from each group.

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Step 2.1.5.2.1

Group the first two terms and the last two terms.

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

Step 2.1.5.2.2

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

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

Step 2.1.5.3

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

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

Step 2.1.6

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

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

Step 2.1.7

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

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Step 2.1.7.1

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

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

Step 2.1.7.2

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 2.1.7.3

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 2.1.8

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

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

Step 2.1.9

Factor by grouping.

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Step 2.1.9.1

Reorder terms.

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

Step 2.1.9.2

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 2.1.9.2.1

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

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

Step 2.1.9.2.2

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

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

Step 2.1.9.2.3

Apply the distributive property.

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

Step 2.1.9.2.4

Multiply $u$ by $1$ .

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

Step 2.1.9.3

Factor out the greatest common factor from each group.

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Step 2.1.9.3.1

Group the first two terms and the last two terms.

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

Step 2.1.9.3.2

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

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

Step 2.1.9.4

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

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

Step 2.1.10

Factor.

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Step 2.1.10.1

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

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

Step 2.1.10.2

Remove unnecessary parentheses.

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

Step 2.2

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

$x^{2} + 25 = 0$

$2 x + 1 = 0$

$x - 1 = 0$

Step 2.3

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

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Step 2.3.1

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

$x^{2} + 25 = 0$

Step 2.3.2

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

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Step 2.3.2.1

Subtract $25$ from both sides of the equation.

$x^{2} = - 25$

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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Step 2.3.2.3.1

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

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

Step 2.3.2.3.2

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

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

Step 2.3.2.3.3

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

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

Step 2.3.2.3.4

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

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

Step 2.3.2.3.5

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

$x = \pm i \cdot 5$

Step 2.3.2.3.6

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

$x = \pm 5 i$

Step 2.3.2.4

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

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Step 2.3.2.4.1

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

$x = 5 i$

Step 2.3.2.4.2

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

$x = - 5 i$

Step 2.3.2.4.3

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

$x = 5 i, - 5 i$

Step 2.4

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

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Step 2.4.1

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

$2 x + 1 = 0$

Step 2.4.2

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

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 2.4.2.2

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

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Step 2.4.2.2.1

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

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

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

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

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.4.2.2.3

Simplify the right side.

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Step 2.4.2.2.3.1

Move the negative in front of the fraction.

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

Step 2.5

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

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Step 2.5.1

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

$x - 1 = 0$

Step 2.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.6

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 3