Precalculus Examples

$f (x) = 8 x^{4} - 72 x^{3} + 144 x^{2}$

Step 1

Set $8 x^{4} - 72 x^{3} + 144 x^{2}$ equal to $0$ .

$8 x^{4} - 72 x^{3} + 144 x^{2} = 0$

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

Factor $8 x^{2}$ out of $8 x^{4} - 72 x^{3} + 144 x^{2}$ .

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Step 2.1.1.1

Factor $8 x^{2}$ out of $8 x^{4}$ .

$8 x^{2} (x^{2}) - 72 x^{3} + 144 x^{2} = 0$

Step 2.1.1.2

Factor $8 x^{2}$ out of $- 72 x^{3}$ .

$8 x^{2} (x^{2}) + 8 x^{2} (- 9 x) + 144 x^{2} = 0$

Step 2.1.1.3

Factor $8 x^{2}$ out of $144 x^{2}$ .

$8 x^{2} (x^{2}) + 8 x^{2} (- 9 x) + 8 x^{2} (18) = 0$

Step 2.1.1.4

Factor $8 x^{2}$ out of $8 x^{2} (x^{2}) + 8 x^{2} (- 9 x)$ .

$8 x^{2} (x^{2} - 9 x) + 8 x^{2} (18) = 0$

Step 2.1.1.5

Factor $8 x^{2}$ out of $8 x^{2} (x^{2} - 9 x) + 8 x^{2} (18)$ .

$8 x^{2} (x^{2} - 9 x + 18) = 0$

Step 2.1.2

Factor.

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Step 2.1.2.1

Factor $x^{2} - 9 x + 18$ using the AC method.

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Step 2.1.2.1.1

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 $18$ and whose sum is $- 9$ .

$- 6, - 3$

Step 2.1.2.1.2

Write the factored form using these integers.

$8 x^{2} ((x - 6) (x - 3)) = 0$

Step 2.1.2.2

Remove unnecessary parentheses.

$8 x^{2} (x - 6) (x - 3) = 0$

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 $0$ .

$x^{2} = 0$

$x - 6 = 0$

$x - 3 = 0$

Step 2.3

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

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Step 2.3.1

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

$x^{2} = 0$

Step 2.3.2

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

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Step 2.3.2.1

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

$x = \pm \sqrt{0}$

Step 2.3.2.2

Simplify $\pm \sqrt{0}$ .

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Step 2.3.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 2.3.2.2.2

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

$x = \pm 0$

Step 2.3.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 2.4

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 2.4.2

Add $6$ to both sides of the equation.

$x = 6$

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

The final solution is all the values that make $8 x^{2} (x - 6) (x - 3) = 0$ true.

$x = 0, 6, 3$

Step 3