Divide. Enter your answer as a single completely simplified rational expression with nonnegative exponents. \[ \frac{2 r^{2}+9 r+7}{r^{2}-49} \div \frac{2 r^{2}+13 r+21}{r^{2}-14 r+49} \]
Solución de inteligencia artificial de Upstudy
Responder
Solución

¡Inicia sesión para desbloquear respuestas gratis!
Una plataforma de aprendizaje en la que confían millones de estudiantes y profesores reales.

Respondido por UpStudy AI y revisado por un tutor profesional
Beyond the Answer
To simplify the expression, we first rewrite the division as multiplication by the reciprocal: \[ \frac{2 r^{2}+9 r+7}{r^{2}-49} \cdot \frac{r^{2}-14 r+49}{2 r^{2}+13 r+21} \] Next, we factor each polynomial. 1. The denominator \(r^2 - 49\) is a difference of squares: \[ r^2 - 49 = (r - 7)(r + 7) \] 2. The numerator \(2r^2 + 9r + 7\) can be factored by finding two numbers that multiply to \(2 \times 7 = 14\) and add to \(9\). The factors are \(2\) and \(7\): \[ 2r^2 + 9r + 7 = (2r + 7)(r + 1) \] 3. The polynomial \(r^2 - 14r + 49\) is also a perfect square: \[ r^2 - 14r + 49 = (r - 7)^2 \] 4. Finally, we factor \(2r^2 + 13r + 21\) by looking for two numbers that multiply to \(2 \times 21 = 42\) and add to \(13\), which are \(6\) and \(7\): \[ 2r^2 + 13r + 21 = (2r + 7)(r + 3) \] Now substituting these factors back into the expression, we get: \[ \frac{(2r + 7)(r + 1)}{(r - 7)(r + 7)} \cdot \frac{(r - 7)^2}{(2r + 7)(r + 3)} \] Now we can cancel the common factors \(2r + 7\) and \(r - 7\): \[ = \frac{(r + 1)(r - 7)}{(r + 7)(r + 3)} \] Finally, so we simplified the expression, we can write the answer as: \[ \frac{(r + 1)(r - 7)}{(r + 7)(r + 3)} \] So the final answer is: \[ \frac{(r + 1)(r - 7)}{(r + 7)(r + 3)} \]
