An 8th-degree polynomial needs 9 terms that involve
x⁸, x⁷, ..., x¹, and x⁰.
x=10 implies that (x-10) is a factor of the polynomial according to the Remainder theorem.
Let the polynomial be of the form
f(x) = a₁x⁸ + a₂x⁷ + a₃x⁶ +a₄x⁵ + a₅x⁴ + a₆x³ + a₇x² + a₈x + a₉
The first few lines of the synthetic division are
10 | a₁ a₂ a₃ a₄ a₅ a₆ a₇ a₈ a₉ ( the first row has 9 coefficients)
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a₁
Answer:
The first row has 9 coefficients.
Answer:
5
Step-by-step explanation:
According to the described rule, we have

We can see the pattern

In other words, for all 

Now,

The slope is -3x i think
just a tip there’s this website (i think there’s an app too?). allied khan academy it’s basically for school and there’s lots of videos of what you’re doing
The number of passwords that would be allowed if users were allowed to reuse the letters and numbers is 30891577600
<h3>How to determine the number of passwords</h3>
The given parameters are:
Password length = 8
Letters = 6
Numbers = 2
There are 26 letters and 10 digits.
Since the letters and the numbers can be repeated, then the number of passwords is:
Count = 26^6 * 10^2
Evaluate
Count = 30891577600
Hence, the number of passwords that would be allowed if users were allowed to reuse the letters and numbers is 30891577600
Read more about combination at:
brainly.com/question/11732255
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