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)
-----------------------------------------
a₁
Answer:
The first row has 9 coefficients.
Answer:
Graph A is correct
Step-by-step explanation:
p(x)= x/10
x= 1, 2, 3, 4
Plug in x values in p(x)
when x=1 , then P(1) = 1/10
When x=2 , then P(2) = 2/10
When x=3 , then P(3) = 3/10
When x=4 , then P(4) = 4/10
In the graph y axis has 2/10 , 4/10 , 6/10...
1/10 lies between 0 and 2/10
3/10 lies between 2/10 and 4/10
Graph A is correct
Answer: D
Step-by-step explanation:
all possible rational zeros are the factors of the last term divided by the coefficient of the first term
so it's (±1, ±3, ±9) / (±1, ±2)
(±1, ±3, ±9) / ±1 = ±1, ±3, ±9
(±1, ±3, ±9) / ±2 = ±1/2, ±3/2, ±9/2
--> ±1, ±3, ±9, ±1/2, ±3/2, ±9/2
Answer: (2, 18)
Step-by-step explanation:
When x=2, 
.
So, it should pass through (2, 18).
