Precalculus Examples

${(z - 3)}^{2} - 3 (z - 3) - 28 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = z - 3$ . Substitute $u$ for all occurrences of $z - 3$ .

$u^{2} - 3 u - 28 = 0$

Step 1.2

Factor $u^{2} - 3 u - 28$ 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 $- 28$ and whose sum is $- 3$ .

$- 7, 4$

Step 1.2.2

Write the factored form using these integers.

$(u - 7) (u + 4) = 0$

Step 1.3

Replace all occurrences of $u$ with $z - 3$ .

$(z - 3 - 7) (z - 3 + 4) = 0$

Step 2

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

$z - 3 - 7 = 0$

$z - 3 + 4 = 0$

Step 3

Set $z - 3 - 7$ equal to $0$ and solve for $z$ .

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Step 3.1

Set $z - 3 - 7$ equal to $0$ .

$z - 3 - 7 = 0$

Step 3.2

Solve $z - 3 - 7 = 0$ for $z$ .

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Step 3.2.1

Subtract $7$ from $- 3$ .

$z - 10 = 0$

Step 3.2.2

Add $10$ to both sides of the equation.

$z = 10$

Step 4

Set $z - 3 + 4$ equal to $0$ and solve for $z$ .

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Step 4.1

Set $z - 3 + 4$ equal to $0$ .

$z - 3 + 4 = 0$

Step 4.2

Solve $z - 3 + 4 = 0$ for $z$ .

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Step 4.2.1

Add $- 3$ and $4$ .

$z + 1 = 0$

Step 4.2.2

Subtract $1$ from both sides of the equation.

$z = - 1$

Step 5

The final solution is all the values that make $(z - 3 - 7) (z - 3 + 4) = 0$ true.

$z = 10, - 1$