Precalculus Examples

$F (x) = 4 x^{4} + 5 x^{3} + 29 x^{2} + 35 x + 7$

Step 1

Set $4 x^{4} + 5 x^{3} + 29 x^{2} + 35 x + 7$ equal to $0$ .

$4 x^{4} + 5 x^{3} + 29 x^{2} + 35 x + 7 = 0$

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.

$5 x^{3} + 35 x + 4 x^{4} + 29 x^{2} + 7 = 0$

Step 2.1.2

Factor $5 x$ out of $5 x^{3} + 35 x$ .

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Step 2.1.2.1

Factor $5 x$ out of $5 x^{3}$ .

$5 x (x^{2}) + 35 x + 4 x^{4} + 29 x^{2} + 7 = 0$

Step 2.1.2.2

Factor $5 x$ out of $35 x$ .

$5 x (x^{2}) + 5 x (7) + 4 x^{4} + 29 x^{2} + 7 = 0$

Step 2.1.2.3

Factor $5 x$ out of $5 x (x^{2}) + 5 x (7)$ .

$5 x (x^{2} + 7) + 4 x^{4} + 29 x^{2} + 7 = 0$

Step 2.1.3

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

$5 x (x^{2} + 7) + 4 {(x^{2})}^{2} + 29 x^{2} + 7 = 0$

Step 2.1.4

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

$5 x (x^{2} + 7) + 4 u^{2} + 29 u + 7 = 0$

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 $a \cdot c = 4 \cdot 7 = 28$ and whose sum is $b = 29$ .

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Step 2.1.5.1.1

Factor $29$ out of $29 u$ .

$5 x (x^{2} + 7) + 4 u^{2} + 29 (u) + 7 = 0$

Step 2.1.5.1.2

Rewrite $29$ as $1$ plus $28$

$5 x (x^{2} + 7) + 4 u^{2} + (1 + 28) u + 7 = 0$

Step 2.1.5.1.3

Apply the distributive property.

$5 x (x^{2} + 7) + 4 u^{2} + 1 u + 28 u + 7 = 0$

Step 2.1.5.1.4

Multiply $u$ by $1$ .

$5 x (x^{2} + 7) + 4 u^{2} + u + 28 u + 7 = 0$

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.

$5 x (x^{2} + 7) + 4 u^{2} + u + 28 u + 7 = 0$

Step 2.1.5.2.2

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

$5 x (x^{2} + 7) + u (4 u + 1) + 7 (4 u + 1) = 0$

Step 2.1.5.3

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

$5 x (x^{2} + 7) + (4 u + 1) (u + 7) = 0$

Step 2.1.6

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

$5 x (x^{2} + 7) + (4 x^{2} + 1) (x^{2} + 7) = 0$

Step 2.1.7

Factor $x^{2} + 7$ out of $5 x (x^{2} + 7) + (4 x^{2} + 1) (x^{2} + 7)$ .

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Step 2.1.7.1

Factor $x^{2} + 7$ out of $5 x (x^{2} + 7)$ .

$(x^{2} + 7) (5 x) + (4 x^{2} + 1) (x^{2} + 7) = 0$

Step 2.1.7.2

Factor $x^{2} + 7$ out of $(4 x^{2} + 1) (x^{2} + 7)$ .

$(x^{2} + 7) (5 x) + (x^{2} + 7) (4 x^{2} + 1) = 0$

Step 2.1.7.3

Factor $x^{2} + 7$ out of $(x^{2} + 7) (5 x) + (x^{2} + 7) (4 x^{2} + 1)$ .

$(x^{2} + 7) (5 x + 4 x^{2} + 1) = 0$

Step 2.1.8

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

$(x^{2} + 7) (5 u + 4 u^{2} + 1) = 0$

Step 2.1.9

Factor by grouping.

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Step 2.1.9.1

Reorder terms.

$(x^{2} + 7) (4 u^{2} + 5 u + 1) = 0$

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 $a \cdot c = 4 \cdot 1 = 4$ and whose sum is $b = 5$ .

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Step 2.1.9.2.1

Factor $5$ out of $5 u$ .

$(x^{2} + 7) (4 u^{2} + 5 (u) + 1) = 0$

Step 2.1.9.2.2

Rewrite $5$ as $1$ plus $4$

$(x^{2} + 7) (4 u^{2} + (1 + 4) u + 1) = 0$

Step 2.1.9.2.3

Apply the distributive property.

$(x^{2} + 7) (4 u^{2} + 1 u + 4 u + 1) = 0$

Step 2.1.9.2.4

Multiply $u$ by $1$ .

$(x^{2} + 7) (4 u^{2} + u + 4 u + 1) = 0$

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.

$(x^{2} + 7) ((4 u^{2} + u) + 4 u + 1) = 0$

Step 2.1.9.3.2

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

$(x^{2} + 7) (u (4 u + 1) + 1 (4 u + 1)) = 0$

Step 2.1.9.4

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

$(x^{2} + 7) ((4 u + 1) (u + 1)) = 0$

Step 2.1.10

Factor.

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Step 2.1.10.1

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

$(x^{2} + 7) ((4 x + 1) (x + 1)) = 0$

Step 2.1.10.2

Remove unnecessary parentheses.

$(x^{2} + 7) (4 x + 1) (x + 1) = 0$

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 $0$ .

$x^{2} + 7 = 0$

$4 x + 1 = 0$

$x + 1 = 0$

Step 2.3

Set $x^{2} + 7$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x^{2} + 7$ equal to $0$ .

$x^{2} + 7 = 0$

Step 2.3.2

Solve $x^{2} + 7 = 0$ for $x$ .

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Step 2.3.2.1

Subtract $7$ from both sides of the equation.

$x^{2} = - 7$

Step 2.3.2.2

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

$x = \pm \sqrt{- 7}$

Step 2.3.2.3

Simplify $\pm \sqrt{- 7}$ .

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Step 2.3.2.3.1

Rewrite $- 7$ as $- 1 (7)$ .

$x = \pm \sqrt{- 1 (7)}$

Step 2.3.2.3.2

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{7}$

Step 2.3.2.3.3

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

$x = \pm i \sqrt{7}$

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.

$x = i \sqrt{7}$

Step 2.3.2.4.2

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

$x = - i \sqrt{7}$

Step 2.3.2.4.3

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

$x = i \sqrt{7}, - i \sqrt{7}$

Step 2.4

Set $4 x + 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $4 x + 1$ equal to $0$ .

$4 x + 1 = 0$

Step 2.4.2

Solve $4 x + 1 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$4 x = - 1$

Step 2.4.2.2

Divide each term in $4 x = - 1$ by $4$ and simplify.

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Step 2.4.2.2.1

Divide each term in $4 x = - 1$ by $4$ .

$\frac{4 x}{4} = \frac{- 1}{4}$

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 $4$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 1}{4}$

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{4}$

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.

$x = - \frac{1}{4}$

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

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.6

The final solution is all the values that make $(x^{2} + 7) (4 x + 1) (x + 1) = 0$ true.

$x = i \sqrt{7}, - i \sqrt{7}, - \frac{1}{4}, - 1$

Step 3