QuestionSimpify: $\frac{x^{2}+x-20}{x^{2}+6 x+5} \div \frac{x^{2}-x-12}{x^{2}-4 x-5}$

Studdy Solution

STEP 1

1. We are asked to simplify a complex rational expression.
2. Division of fractions can be converted to multiplication by the reciprocal.
3. Factoring polynomials will simplify the expression.

STEP 2

1. Rewrite the division as multiplication by the reciprocal.
2. Factor all polynomials in the expression.
3. Simplify the expression by canceling common factors.

STEP 3

Rewrite the division as multiplication by the reciprocal:
$\frac{x^{2}+x-20}{x^{2}+6x+5} \div \frac{x^{2}-x-12}{x^{2}-4x-5} = \frac{x^{2}+x-20}{x^{2}+6x+5} \times \frac{x^{2}-4x-5}{x^{2}-x-12}$

STEP 4

Factor each polynomial:
- $x^{2}+x-20$ factors to $(x+5)(x-4)$ . - $x^{2}+6x+5$ factors to $(x+5)(x+1)$ . - $x^{2}-x-12$ factors to $(x-4)(x+3)$ . - $x^{2}-4x-5$ factors to $(x-5)(x+1)$ .

STEP 5

Substitute the factored forms into the expression:
$\frac{(x+5)(x-4)}{(x+5)(x+1)} \times \frac{(x-5)(x+1)}{(x-4)(x+3)}$
Cancel common factors:
- Cancel $(x+5)$ from the numerator and denominator. - Cancel $(x+1)$ from the numerator and denominator. - Cancel $(x-4)$ from the numerator and denominator.
The simplified expression is:
$\frac{x-5}{x+3}$

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