Pre-Algebra Examples

$- x^{2} + 28 x + 29 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $- 1$ out of $- x^{2} + 28 x + 29$ .

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Step 1.1.1

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 28 x + 29 = 0$

Step 1.1.2

Factor $- 1$ out of $28 x$ .

$- (x^{2}) - (- 28 x) + 29 = 0$

Step 1.1.3

Rewrite $29$ as $- 1 (- 29)$ .

$- (x^{2}) - (- 28 x) - 1 \cdot - 29 = 0$

Step 1.1.4

Factor $- 1$ out of $- (x^{2}) - (- 28 x)$ .

$- (x^{2} - 28 x) - 1 \cdot - 29 = 0$

Step 1.1.5

Factor $- 1$ out of $- (x^{2} - 28 x) - 1 (- 29)$ .

$- (x^{2} - 28 x - 29) = 0$

Step 1.2

Factor.

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Step 1.2.1

Factor $x^{2} - 28 x - 29$ using the AC method.

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Step 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 $- 29$ and whose sum is $- 28$ .

$- 29, 1$

Step 1.2.1.2

Write the factored form using these integers.

$- ((x - 29) (x + 1)) = 0$

Step 1.2.2

Remove unnecessary parentheses.

$- (x - 29) (x + 1) = 0$

Step 2

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

$x - 29 = 0$

$x + 1 = 0$

Step 3

Set $x - 29$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x - 29$ equal to $0$ .

$x - 29 = 0$

Step 3.2

Add $29$ to both sides of the equation.

$x = 29$

Step 4

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

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Step 4.1

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

$x + 1 = 0$

Step 4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

The final solution is all the values that make $- (x - 29) (x + 1) = 0$ true.

$x = 29, - 1$