Prove:
Using mathemetical induction:
P(n) = 
for n=1
P(n) = 
= 6
It is divisible by 2 and 3
Now, for n=k, 
P(k) = 
Assuming P(k) is divisible by 2 and 3:
Now, for n=k+1:
P(k+1) = 
P(k+1) = 
P(k+1) = 
Since, we assumed that P(k) is divisible by 2 and 3, therefore, P(k+1) is also
divisible by 2 and 3.
Hence, by mathematical induction, P(n) = is divisible by 2 and 3 for all positive integer n.
Answer:
its B
Step-by-step explanation:
The polynomial whose zeroes are 1, 2 and 3 is given by,
(x-1)(x-2)(x-3) = 0
(x-1)[x²-3x-2x+6]=0
(x-1)[x²-5x+6]=0
x[x²-5x+6] -1[x²-5x+6]=0
x³-5x²+6x-x²+5x-6=0
x³-6x²+11x-6=0
Therefore, the required polynomial is,
x³ - 6x² + 11x -6 = 0
I believe you would multiply 6 and 10 and get the area of 60, thats all i can think of
