Answer:
YES
NO
NO
Step-by-step explanation:
The given polynomial is: 
(x - a) is a factor of a polynomial iff x = a is a solution to the polynomial.
To check if (x - 5) is a factor of the polynomial f(x), we substitute x = 5 and check if it satisfies the equation.
∴ f(5) = 5³ + 4(5)² - 25(5) - 100
= 125 + 100 - 125 - 100
= 225 - 225
= 0
We see, x = 5 satisfies f(x). So, (x - 5) is a factor to the polynomial.
Now, to check (x + 2) is a factor.
i.e., to check x = - 2 satisfies f(x) or not.
f(-2) = (-2)³ + 4(-2)² - 25(-2) - 100
= -8 + 16 + 50 - 100
= -108 + 66
≠ 0
Therefore, (x + 2) is not a factor of f(x).
To check (x - 4) is a factor.
∴ f(4) = 4³ + 4(4)² - 25(4) - 100
= 64 + 64 - 100 - 100
= 128 - 200
≠ 0
Therefore, (x - 4) is not a factor of f(x).
Answer:
A. 4x²+20x+25
Step-by-step explanation:
Because of PEMDAS, you calculate the value inside of the parenthesis first, so you expand [2x+1-(4x+6)]:
2x+1-4x-6 = -2x-5.
Then apply the exponent:
(-2x-5)² = (-2x-5)(-2x-5) = 4x²+10x+10x+25 = 4x²+20x+25
It might be mmmm idk I’m too dome
The first term, a, is 2. The common ratio, r, is 4. Thus,
a_(n+1) = 2(4)^(n).
Check: What's the first term? Let n=1. Then we get 2(4)^1, or 8. Is that correct? No.
Try this instead:
a_(n) = a_0*4^(n-1). Is this correct? Seeking the first term (n=1), does this formula produce 2? 2*4^0 = 2*1 = 2. YES.
The desired explicit formula is a_(n) = a_0*4^(n-1), where n begins at 1.