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dem82 [27]
3 years ago
11

A prime polynomial is an irreducible polynomial with integer coefficients that cannot be factored into polynomials of lower degr

ee over the real number system. Which trinomial is prime?
A.) x2 – 7x – 18
B.) x2 – 9x – 18
C.)x2 + 3x – 18
D.)x2 + 17x – 18
Mathematics
2 answers:
ELEN [110]3 years ago
8 0

your answer is


b. x²-9x-18


have a good day :) :) :)

SpyIntel [72]3 years ago
4 0
The answer is B.) x^2 - 9x - 18.
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Which problem can we solve with 1/2 / 7?
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Step-by-step explanation:

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2 years ago
Need an answer as soon as possible.
Elza [17]
Part A

Answer: The common ratio is -2

-----------------------------------

Explanation: 

To get the common ratio r, we divide any term by the previous one

One example:
r = common ratio
r = (second term)/(first term)
r = (-2)/(1)
r = -2

Another example:
r = common ratio
r = (third term)/(second term)
r = (4)/(-2)
r = -2
and we get the same common ratio every time

Side Note: each term is multiplied by -2 to get the next term

============================================================
Part B

Answer:
The rule for the sequence is 
a(n) = (-2)^(n-1)
where n starts at n = 1

-----------------------------------

Explanation:

Recall that any geometric sequence has the nth term
a(n) = a*(r)^(n-1)
where the 'a' on the right side is the first term and r is the common ratio

The first term given to use is a = 1 and the common ratio found in part A above was r = -2
So,
a(n) = a*(r)^(n-1)
a(n) = 1*(-2)^(n-1)
a(n) = (-2)^(n-1)

============================================================
Part C

Answer: The next three terms are 16, -32, 64

-----------------------------------

Explanation:

We can simply multiply each previous term by -2 to get the next term. Do this three times to generate the next three terms

-8*(-2) = 16
16*(-2) = -32
-32*(-2) = 64

showing that the next three terms are 16, -32, and 64

An alternative is to use the formula found in part B

Plug in n = 5 to find the fifth term
a(n) = (-2)^(n-1)
a(5) = (-2)^(5-1)
a(5) = (-2)^(4)
a(5) = 16 .... which matches with what we got earlier

Then plug in n = 6
a(n) = (-2)^(n-1)
a(6) = (-2)^(6-1)
a(6) = (-2)^(5)
a(6) = -32 .... which matches with what we got earlier

Then plug in n = 7
a(n) = (-2)^(n-1)
a(7) = (-2)^(7-1)
a(7) = (-2)^(6)
a(7) = 64 .... which matches with what we got earlier

while the second method takes a bit more work, its handy for when you want to find terms beyond the given sequence (eg: the 28th term)
6 0
3 years ago
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