Precalculus Examples

$\frac{x^{2} + 5 x + 6}{(x^{2} - 1) (x - 7)} = 1$

Step 1

Simplify $\frac{x^{2} + 5 x + 6}{(x^{2} - 1) (x - 7)}$ .

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Step 1.1

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

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Step 1.1.1

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 $6$ and whose sum is $5$ .

$2, 3$

Step 1.1.2

Write the factored form using these integers.

$\frac{(x + 2) (x + 3)}{(x^{2} - 1) (x - 7)} = 1$

Step 1.2

Simplify the denominator.

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Step 1.2.1

Rewrite $1$ as $1^{2}$ .

$\frac{(x + 2) (x + 3)}{(x^{2} - 1^{2}) (x - 7)} = 1$

Step 1.2.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\frac{(x + 2) (x + 3)}{(x + 1) (x - 1) (x - 7)} = 1$

Step 2

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx - 0.81884187, 0.14072605, 8.67811581$

Step 3