Select the two binomials that are factors of this trinomial.
2 answers:
To split this trinomial into two binomials, let's try and find two numbers which add to 6 and multiply to 8. To do this, we can list all the factors of 8 and then choose which factors also add to 6.
Factors of 8: (1, 8), (2, 4)
1 + 8 = 9, meaning that 1 and 8 are not the factors we are looking for. However, 2 and 4 do add to 6. By combining these numbers which an x (so that we can produce the term at the front of the trinomial), we find the binomials:
and
The answer is x + 2 and x + 4.
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Step-by-step explanation:
Answer:
Step-by-step explanation:
Lets divide it in cases, then sum everything
Case (1): All 5 numbers are different
In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.
The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.
Case (2): 4 numbers are different
We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4
We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.
The total cardinality of this case is
Case (3): 3 numbers are different
As we did before, we pick 3 elements from a set of n. The amount of possibilities is
Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of
Case (4): 2 numbers are different
We pick 2 numbers from a set of n, with a total of possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.
The total amount of possibilities for this case is
Case (5): All numbers are the same
This is easy, he have as many possibilities as numbers the set has. In other words, n
Conclussion
By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is
I hope that works for you!