Finite Math Examples

${(x^{2} - 6 x)}^{2} - 2 (x^{2} - 6 x) - 35 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = x^{2} - 6 x$ . Substitute $u$ for all occurrences of $x^{2} - 6 x$ .

$u^{2} - 2 u - 35 = 0$

Step 1.2

Factor $u^{2} - 2 u - 35$ using the AC method.

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Step 1.2.1

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 $- 35$ and whose sum is $- 2$ .

$- 7, 5$

Step 1.2.2

Write the factored form using these integers.

$(u - 7) (u + 5) = 0$

Step 1.3

Replace all occurrences of $u$ with $x^{2} - 6 x$ .

$(x^{2} - 6 x - 7) (x^{2} - 6 x + 5) = 0$

Step 2

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

$x^{2} - 6 x - 7 = 0$

$x^{2} - 6 x + 5 = 0$

Step 3

Set $x^{2} - 6 x - 7$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x^{2} - 6 x - 7$ equal to $0$ .

$x^{2} - 6 x - 7 = 0$

Step 3.2

Solve $x^{2} - 6 x - 7 = 0$ for $x$ .

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Step 3.2.1

Factor $x^{2} - 6 x - 7$ using the AC method.

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Step 3.2.1.1

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 $- 7$ and whose sum is $- 6$ .

$- 7, 1$

Step 3.2.1.2

Write the factored form using these integers.

$(x - 7) (x + 1) = 0$

Step 3.2.2

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

$x - 7 = 0$

$x + 1 = 0$

Step 3.2.3

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 3.2.3.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 3.2.3.2

Add $7$ to both sides of the equation.

$x = 7$

Step 3.2.4

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

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Step 3.2.4.1

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

$x + 1 = 0$

Step 3.2.4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.2.5

The final solution is all the values that make $(x - 7) (x + 1) = 0$ true.

$x = 7, - 1$

Step 4

Set $x^{2} - 6 x + 5$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x^{2} - 6 x + 5$ equal to $0$ .

$x^{2} - 6 x + 5 = 0$

Step 4.2

Solve $x^{2} - 6 x + 5 = 0$ for $x$ .

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Step 4.2.1

Factor $x^{2} - 6 x + 5$ using the AC method.

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Step 4.2.1.1

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 $5$ and whose sum is $- 6$ .

$- 5, - 1$

Step 4.2.1.2

Write the factored form using these integers.

$(x - 5) (x - 1) = 0$

Step 4.2.2

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

$x - 5 = 0$

$x - 1 = 0$

Step 4.2.3

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

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Step 4.2.3.1

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

$x - 5 = 0$

Step 4.2.3.2

Add $5$ to both sides of the equation.

$x = 5$

Step 4.2.4

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 4.2.4.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 4.2.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 4.2.5

The final solution is all the values that make $(x - 5) (x - 1) = 0$ true.

$x = 5, 1$

Step 5

The final solution is all the values that make $(x^{2} - 6 x - 7) (x^{2} - 6 x + 5) = 0$ true.

$x = 7, - 1, 5, 1$