Finite Math Examples

${(x + 5)}^{2} - 12 (x + 5) + 32 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = x + 5$ . Substitute $u$ for all occurrences of $x + 5$ .

$u^{2} - 12 u + 32 = 0$

Step 1.2

Factor $u^{2} - 12 u + 32$ 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 $32$ and whose sum is $- 12$ .

$- 8, - 4$

Step 1.2.2

Write the factored form using these integers.

$(u - 8) (u - 4) = 0$

Step 1.3

Replace all occurrences of $u$ with $x + 5$ .

$(x + 5 - 8) (x + 5 - 4) = 0$

Step 2

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

$x + 5 - 8 = 0$

$x + 5 - 4 = 0$

Step 3

Set $x + 5 - 8$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x + 5 - 8$ equal to $0$ .

$x + 5 - 8 = 0$

Step 3.2

Solve $x + 5 - 8 = 0$ for $x$ .

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Step 3.2.1

Subtract $8$ from $5$ .

$x - 3 = 0$

Step 3.2.2

Add $3$ to both sides of the equation.

$x = 3$

Step 4

Set $x + 5 - 4$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x + 5 - 4$ equal to $0$ .

$x + 5 - 4 = 0$

Step 4.2

Solve $x + 5 - 4 = 0$ for $x$ .

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Step 4.2.1

Subtract $4$ from $5$ .

$x + 1 = 0$

Step 4.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

The final solution is all the values that make $(x + 5 - 8) (x + 5 - 4) = 0$ true.

$x = 3, - 1$

Step 6