Precalculus Examples

${(m^{2} - 1)}^{2} - 11 (m^{2} - 1) + 24 = 0$

Step 1

Let $u = m^{2} - 1$ . Substitute $u$ for all occurrences of $m^{2} - 1$ .

$u^{2} - 11 u + 24 = 0$

Step 2

Factor $u^{2} - 11 u + 24$ using the AC method.

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Step 2.1

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 $24$ and whose sum is $- 11$ .

$- 8, - 3$

Step 2.2

Write the factored form using these integers.

$(u - 8) (u - 3) = 0$

Step 3

Replace all occurrences of $u$ with $m^{2} - 1$ .

$(m^{2} - 1 - 8) (m^{2} - 1 - 3) = 0$

Step 4

Simplify.

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Step 4.1

Subtract $8$ from $- 1$ .

$(m^{2} - 9) (m^{2} - 1 - 3) = 0$

Step 4.2

Subtract $3$ from $- 1$ .

$(m^{2} - 9) (m^{2} - 4) = 0$

Step 5

Rewrite $9$ as $3^{2}$ .

$(m^{2} - 3^{2}) (m^{2} - 4) = 0$

Step 6

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

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

Step 7

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

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

Step 8

Factor.

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Step 8.1

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

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

Step 8.2

Remove unnecessary parentheses.

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

Step 9

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

$m + 3 = 0$

$m - 3 = 0$

$m + 2 = 0$

$m - 2 = 0$

Step 10

Set $m + 3$ equal to $0$ and solve for $m$ .

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Step 10.1

Set $m + 3$ equal to $0$ .

$m + 3 = 0$

Step 10.2

Subtract $3$ from both sides of the equation.

$m = - 3$

Step 11

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

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Step 11.1

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

$m - 3 = 0$

Step 11.2

Add $3$ to both sides of the equation.

$m = 3$

Step 12

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

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Step 12.1

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

$m + 2 = 0$

Step 12.2

Subtract $2$ from both sides of the equation.

$m = - 2$

Step 13

Set $m - 2$ equal to $0$ and solve for $m$ .

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Step 13.1

Set $m - 2$ equal to $0$ .

$m - 2 = 0$

Step 13.2

Add $2$ to both sides of the equation.

$m = 2$

Step 14

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

$m = - 3, 3, - 2, 2$