Precalculus Examples

$f (x) = 25 x^{4} + 126 x^{3} + 180 x^{2} + 882 x + 35$

Step 1

Set $25 x^{4} + 126 x^{3} + 180 x^{2} + 882 x + 35$ equal to $0$ .

$25 x^{4} + 126 x^{3} + 180 x^{2} + 882 x + 35 = 0$

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.

$126 x^{3} + 882 x + 25 x^{4} + 180 x^{2} + 35 = 0$

Step 2.1.2

Factor $126 x$ out of $126 x^{3} + 882 x$ .

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Step 2.1.2.1

Factor $126 x$ out of $126 x^{3}$ .

$126 x (x^{2}) + 882 x + 25 x^{4} + 180 x^{2} + 35 = 0$

Step 2.1.2.2

Factor $126 x$ out of $882 x$ .

$126 x (x^{2}) + 126 x (7) + 25 x^{4} + 180 x^{2} + 35 = 0$

Step 2.1.2.3

Factor $126 x$ out of $126 x (x^{2}) + 126 x (7)$ .

$126 x (x^{2} + 7) + 25 x^{4} + 180 x^{2} + 35 = 0$

Step 2.1.3

Factor $5$ out of $25 x^{4} + 180 x^{2} + 35$ .

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Step 2.1.3.1

Factor $5$ out of $25 x^{4}$ .

$126 x (x^{2} + 7) + 5 (5 x^{4}) + 180 x^{2} + 35 = 0$

Step 2.1.3.2

Factor $5$ out of $180 x^{2}$ .

$126 x (x^{2} + 7) + 5 (5 x^{4}) + 5 (36 x^{2}) + 35 = 0$

Step 2.1.3.3

Factor $5$ out of $35$ .

$126 x (x^{2} + 7) + 5 (5 x^{4}) + 5 (36 x^{2}) + 5 (7) = 0$

Step 2.1.3.4

Factor $5$ out of $5 (5 x^{4}) + 5 (36 x^{2})$ .

$126 x (x^{2} + 7) + 5 (5 x^{4} + 36 x^{2}) + 5 (7) = 0$

Step 2.1.3.5

Factor $5$ out of $5 (5 x^{4} + 36 x^{2}) + 5 (7)$ .

$126 x (x^{2} + 7) + 5 (5 x^{4} + 36 x^{2} + 7) = 0$

Step 2.1.4

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

$126 x (x^{2} + 7) + 5 (5 {(x^{2})}^{2} + 36 x^{2} + 7) = 0$

Step 2.1.5

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

$126 x (x^{2} + 7) + 5 (5 u^{2} + 36 u + 7) = 0$

Step 2.1.6

Factor by grouping.

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Step 2.1.6.1

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 $a \cdot c = 5 \cdot 7 = 35$ and whose sum is $b = 36$ .

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Step 2.1.6.1.1

Factor $36$ out of $36 u$ .

$126 x (x^{2} + 7) + 5 (5 u^{2} + 36 (u) + 7) = 0$

Step 2.1.6.1.2

Rewrite $36$ as $1$ plus $35$

$126 x (x^{2} + 7) + 5 (5 u^{2} + (1 + 35) u + 7) = 0$

Step 2.1.6.1.3

Apply the distributive property.

$126 x (x^{2} + 7) + 5 (5 u^{2} + 1 u + 35 u + 7) = 0$

Step 2.1.6.1.4

Multiply $u$ by $1$ .

$126 x (x^{2} + 7) + 5 (5 u^{2} + u + 35 u + 7) = 0$

Step 2.1.6.2

Factor out the greatest common factor from each group.

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Step 2.1.6.2.1

Group the first two terms and the last two terms.

$126 x (x^{2} + 7) + 5 ((5 u^{2} + u) + 35 u + 7) = 0$

Step 2.1.6.2.2

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

$126 x (x^{2} + 7) + 5 (u (5 u + 1) + 7 (5 u + 1)) = 0$

Step 2.1.6.3

Factor the polynomial by factoring out the greatest common factor, $5 u + 1$ .

$126 x (x^{2} + 7) + 5 ((5 u + 1) (u + 7)) = 0$

Step 2.1.7

Factor.

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Step 2.1.7.1

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

$126 x (x^{2} + 7) + 5 ((5 x^{2} + 1) (x^{2} + 7)) = 0$

Step 2.1.7.2

Remove unnecessary parentheses.

$126 x (x^{2} + 7) + 5 (5 x^{2} + 1) (x^{2} + 7) = 0$

Step 2.1.8

Factor $x^{2} + 7$ out of $126 x (x^{2} + 7) + 5 (5 x^{2} + 1) (x^{2} + 7)$ .

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Step 2.1.8.1

Factor $x^{2} + 7$ out of $126 x (x^{2} + 7)$ .

$(x^{2} + 7) (126 x) + 5 (5 x^{2} + 1) (x^{2} + 7) = 0$

Step 2.1.8.2

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

$(x^{2} + 7) (126 x) + (x^{2} + 7) (5 (5 x^{2} + 1)) = 0$

Step 2.1.8.3

Factor $x^{2} + 7$ out of $(x^{2} + 7) (126 x) + (x^{2} + 7) (5 (5 x^{2} + 1))$ .

$(x^{2} + 7) (126 x + 5 (5 x^{2} + 1)) = 0$

Step 2.1.9

Apply the distributive property.

$(x^{2} + 7) (126 x + 5 (5 x^{2}) + 5 \cdot 1) = 0$

Step 2.1.10

Multiply $5$ by $5$ .

$(x^{2} + 7) (126 x + 25 x^{2} + 5 \cdot 1) = 0$

Step 2.1.11

Multiply $5$ by $1$ .

$(x^{2} + 7) (126 x + 25 x^{2} + 5) = 0$

Step 2.1.12

Reorder terms.

$(x^{2} + 7) (25 x^{2} + 126 x + 5) = 0$

Step 2.1.13

Factor.

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Step 2.1.13.1

Factor by grouping.

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Step 2.1.13.1.1

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 $a \cdot c = 25 \cdot 5 = 125$ and whose sum is $b = 126$ .

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Step 2.1.13.1.1.1

Factor $126$ out of $126 x$ .

$(x^{2} + 7) (25 x^{2} + 126 (x) + 5) = 0$

Step 2.1.13.1.1.2

Rewrite $126$ as $1$ plus $125$

$(x^{2} + 7) (25 x^{2} + (1 + 125) x + 5) = 0$

Step 2.1.13.1.1.3

Apply the distributive property.

$(x^{2} + 7) (25 x^{2} + 1 x + 125 x + 5) = 0$

Step 2.1.13.1.1.4

Multiply $x$ by $1$ .

$(x^{2} + 7) (25 x^{2} + x + 125 x + 5) = 0$

Step 2.1.13.1.2

Factor out the greatest common factor from each group.

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Step 2.1.13.1.2.1

Group the first two terms and the last two terms.

$(x^{2} + 7) ((25 x^{2} + x) + 125 x + 5) = 0$

Step 2.1.13.1.2.2

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

$(x^{2} + 7) (x (25 x + 1) + 5 (25 x + 1)) = 0$

Step 2.1.13.1.3

Factor the polynomial by factoring out the greatest common factor, $25 x + 1$ .

$(x^{2} + 7) ((25 x + 1) (x + 5)) = 0$

Step 2.1.13.2

Remove unnecessary parentheses.

$(x^{2} + 7) (25 x + 1) (x + 5) = 0$

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 $0$ .

$x^{2} + 7 = 0$

$25 x + 1 = 0$

$x + 5 = 0$

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 $25 x + 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $25 x + 1$ equal to $0$ .

$25 x + 1 = 0$

Step 2.4.2

Solve $25 x + 1 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$25 x = - 1$

Step 2.4.2.2

Divide each term in $25 x = - 1$ by $25$ and simplify.

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Step 2.4.2.2.1

Divide each term in $25 x = - 1$ by $25$ .

$\frac{25 x}{25} = \frac{- 1}{25}$

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 $25$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{25 x}{25} = \frac{- 1}{25}$

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{25}$

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.

$x = - \frac{1}{25}$

Step 2.5

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

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Step 2.5.1

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

$x + 5 = 0$

Step 2.5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.6

The final solution is all the values that make $(x^{2} + 7) (25 x + 1) (x + 5) = 0$ true.

$x = i \sqrt{7}, - i \sqrt{7}, - \frac{1}{25}, - 5$

Step 3