Answer:
3x³ + x² + x + 1
= 3x³ - 3x² + 4x² - 4x + 5x - 5 + 6
= 3x²(x - 1) + 4x(x - 1) + 5(x - 1) + 6
= (x - 1)(3x² + 4x + 5) + 6
but 3x³ + x² + x + 1 = (x - 1).Q(x) + R
=> Q(x) = 3x² + 4x + 5
R = 6
Answer:
3840
Step-by-step explanation:
Using the fundamental counting principle
The restaurant is offering five kinds of soup, three kinds of salad, four kinds of appetizers, six entrees, five desserts, and four beverages.
There are 5 independent events happening (5 prixe dinners)
(1) soup or salad (not both)
We have five kinds of soup, three kinds of salad
soup or salad (not both), = 5 + 3
= 8
(2) appetizer = 4
(3) entrees = 6
(4) dessert = 5
(5) Beverage = 4
Hence,
8 × 4 × 6 × 5 × 4
= 3840
The number of different prix fixe dinners are possible is 3840
