Precalculus Examples

$f (x) = 9 x^{4} - 9 x^{3} - 58 x^{2} + 4 x + 24$

Step 1

Set $9 x^{4} - 9 x^{3} - 58 x^{2} + 4 x + 24$ equal to $0$ .

$9 x^{4} - 9 x^{3} - 58 x^{2} + 4 x + 24 = 0$

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.

$- 9 x^{3} + 4 x + 9 x^{4} - 58 x^{2} + 24 = 0$

Step 2.1.2

Factor $- x$ out of $- 9 x^{3} + 4 x$ .

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Step 2.1.2.1

Factor $- x$ out of $- 9 x^{3}$ .

$- x (9 x^{2}) + 4 x + 9 x^{4} - 58 x^{2} + 24 = 0$

Step 2.1.2.2

Factor $- x$ out of $4 x$ .

$- x (9 x^{2}) - x \cdot - 4 + 9 x^{4} - 58 x^{2} + 24 = 0$

Step 2.1.2.3

Factor $- x$ out of $- x (9 x^{2}) - x (- 4)$ .

$- x (9 x^{2} - 4) + 9 x^{4} - 58 x^{2} + 24 = 0$

Step 2.1.3

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$- x ({(3 x)}^{2} - 4) + 9 x^{4} - 58 x^{2} + 24 = 0$

Step 2.1.4

Rewrite $4$ as $2^{2}$ .

$- x ({(3 x)}^{2} - 2^{2}) + 9 x^{4} - 58 x^{2} + 24 = 0$

Step 2.1.5

Factor.

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Step 2.1.5.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 2$ .

$- x ((3 x + 2) (3 x - 2)) + 9 x^{4} - 58 x^{2} + 24 = 0$

Step 2.1.5.2

Remove unnecessary parentheses.

$- x (3 x + 2) (3 x - 2) + 9 x^{4} - 58 x^{2} + 24 = 0$

Step 2.1.6

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

$- x (3 x + 2) (3 x - 2) + 9 {(x^{2})}^{2} - 58 x^{2} + 24 = 0$

Step 2.1.7

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

$- x (3 x + 2) (3 x - 2) + 9 u^{2} - 58 u + 24 = 0$

Step 2.1.8

Factor by grouping.

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Step 2.1.8.1

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 $a \cdot c = 9 \cdot 24 = 216$ and whose sum is $b = - 58$ .

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Step 2.1.8.1.1

Factor $- 58$ out of $- 58 u$ .

$- x (3 x + 2) (3 x - 2) + 9 u^{2} - 58 u + 24 = 0$

Step 2.1.8.1.2

Rewrite $- 58$ as $- 4$ plus $- 54$

$- x (3 x + 2) (3 x - 2) + 9 u^{2} + (- 4 - 54) u + 24 = 0$

Step 2.1.8.1.3

Apply the distributive property.

$- x (3 x + 2) (3 x - 2) + 9 u^{2} - 4 u - 54 u + 24 = 0$

Step 2.1.8.2

Factor out the greatest common factor from each group.

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Step 2.1.8.2.1

Group the first two terms and the last two terms.

$- x (3 x + 2) (3 x - 2) + 9 u^{2} - 4 u - 54 u + 24 = 0$

Step 2.1.8.2.2

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

$- x (3 x + 2) (3 x - 2) + u (9 u - 4) - 6 (9 u - 4) = 0$

Step 2.1.8.3

Factor the polynomial by factoring out the greatest common factor, $9 u - 4$ .

$- x (3 x + 2) (3 x - 2) + (9 u - 4) (u - 6) = 0$

Step 2.1.9

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

$- x (3 x + 2) (3 x - 2) + (9 x^{2} - 4) (x^{2} - 6) = 0$

Step 2.1.10

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$- x (3 x + 2) (3 x - 2) + ({(3 x)}^{2} - 4) (x^{2} - 6) = 0$

Step 2.1.11

Rewrite $4$ as $2^{2}$ .

$- x (3 x + 2) (3 x - 2) + ({(3 x)}^{2} - 2^{2}) (x^{2} - 6) = 0$

Step 2.1.12

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 2$ .

$- x (3 x + 2) (3 x - 2) + (3 x + 2) (3 x - 2) (x^{2} - 6) = 0$

Step 2.1.13

Factor $(3 x + 2) (3 x - 2)$ out of $- x (3 x + 2) (3 x - 2) + (3 x + 2) (3 x - 2) (x^{2} - 6)$ .

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Step 2.1.13.1

Factor $(3 x + 2) (3 x - 2)$ out of $- x (3 x + 2) (3 x - 2)$ .

$(3 x + 2) (3 x - 2) (- x) + (3 x + 2) (3 x - 2) (x^{2} - 6) = 0$

Step 2.1.13.2

Factor $(3 x + 2) (3 x - 2)$ out of $(3 x + 2) (3 x - 2) (- x) + (3 x + 2) (3 x - 2) (x^{2} - 6)$ .

$(3 x + 2) (3 x - 2) (- x + x^{2} - 6) = 0$

Step 2.1.14

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

$(3 x + 2) (3 x - 2) (- u + u^{2} - 6) = 0$

Step 2.1.15

Factor $- u + u^{2} - 6$ using the AC method.

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Step 2.1.15.1

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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Step 2.1.15.2

Write the factored form using these integers.

$(3 x + 2) (3 x - 2) ((u - 3) (u + 2)) = 0$

Step 2.1.16

Factor.

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Step 2.1.16.1

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

$(3 x + 2) (3 x - 2) ((x - 3) (x + 2)) = 0$

Step 2.1.16.2

Remove unnecessary parentheses.

$(3 x + 2) (3 x - 2) (x - 3) (x + 2) = 0$

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 $0$ .

$3 x + 2 = 0$

$3 x - 2 = 0$

$x - 3 = 0$

$x + 2 = 0$

Step 2.3

Set $3 x + 2$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $3 x + 2$ equal to $0$ .

$3 x + 2 = 0$

Step 2.3.2

Solve $3 x + 2 = 0$ for $x$ .

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Step 2.3.2.1

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Step 2.3.2.2

Divide each term in $3 x = - 2$ by $3$ and simplify.

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Step 2.3.2.2.1

Divide each term in $3 x = - 2$ by $3$ .

$\frac{3 x}{3} = \frac{- 2}{3}$

Step 2.3.2.2.2

Simplify the left side.

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Step 2.3.2.2.2.1

Cancel the common factor of $3$ .

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Step 2.3.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 2}{3}$

Step 2.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 2}{3}$

Step 2.3.2.2.3

Simplify the right side.

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Step 2.3.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

Step 2.4

Set $3 x - 2$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Step 2.4.2

Solve $3 x - 2 = 0$ for $x$ .

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Step 2.4.2.1

Add $2$ to both sides of the equation.

$3 x = 2$

Step 2.4.2.2

Divide each term in $3 x = 2$ by $3$ and simplify.

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Step 2.4.2.2.1

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

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 $3$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Step 2.5

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 2.5.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.6

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

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Step 2.6.1

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

$x + 2 = 0$

Step 2.6.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.7

The final solution is all the values that make $(3 x + 2) (3 x - 2) (x - 3) (x + 2) = 0$ true.

$x = - \frac{2}{3}, \frac{2}{3}, 3, - 2$

Step 3