Question

For any three consecutive numbers, prove algebraically that the largest number and the smallest number are factors of the number that is one less than the square of the middle number.

Answer 1

To prove algebraically that the largest and smallest numbers are factors of the number one less than the square of the middle number, use algebraic expressions and properties.

Step-by-step explanation:

To prove algebraically that the largest number and the smallest number are factors of the number that is one less than the square of the middle number, we can use algebraic expressions and properties.

1. Let the three consecutive numbers be n-1, n, and n+1.
2. The number that is one less than the square of the middle number is (n^2 - 1).
3. We can then show that (n-1) and (n+1) are factors of (n^2 - 1) by performing polynomial division or using the difference of squares formula.

For example, if we let n = 2, the three consecutive numbers are 1, 2, and 3. The middle number squared is 2^2 = 4, and one less than 4 is 3, which is divisible by both 1 and 3.

Answer 2

"For any three consecutive numbers": x,y,z:
x
y=x+1
z=x+2

"prove algebraically that the largest number and the smallest number are factors"
x*z=

"of the number that is one less than the square of the middle number"

=y²-1

x*z=y²-1
x*(x+2)=(x+1)²-1
x²+2x=x²+2x+1-1
x²+2x=x²+2x

both sides of the equation are the same, so the statement is correct