EDGE Math 8

(Continued from the previous slide)

b. Let P be the perimeter of the rectangular picture frame. The perimeter of a rectangle is equal to twice the sum of the length and the width.

P = 2 (\frac{6 p}{p^{2} + 5 p + 6} + \frac{3 p + 6}{p + 3})

= 2 [\frac{6 p}{(p + 3) (p + 2)} + \frac{3 (p + 2)}{p + 3}]

= 2 [\frac{6 p}{(p + 3) (p + 2)} + \frac{3 (p + 2) (p + 2)}{(p + 3) (p + 2)}]

= 2 [\frac{6 p + 3 (p + 2) (p + 2)}{(p + 3) (p + 2)}]

= 2 [\frac{6 p + 3 (p^{2} + 4 p + 4)}{(p + 3) (p + 2)}]

= 2 [\frac{6 p + 3 p^{2} + 12 p + 12}{(p + 3) (p + 2)}]

= 2 [\frac{3 p^{2} + 18 p + 12)}{(p + 3) (p + 2)}]

= 2 [\frac{3 (p^{2} + 6 p + 4)}{(p + 3) (p + 2)}]

= \frac{6 (p^{2} + 6 p + 4)}{(p + 3) (p + 2)}

Therefore, the perimeter of the picture frame is $\frac{6 (p^{2} + 6 p + 4)}{(p + 3) (p + 2)}$ cm.