The sum of the squares of 2 consecutive negative integers is 41. what are the numbers?
which of the following equations is the result of using the factoring method to solve the problem?
(n - 5)(n - 4) = 0
(n - 5)(n + 4) = 0
(n + 5)(n - 4) = 0
(n + 5)(n + 4) = 0

Factors are (n + 5)(n - 4) = 0 .

Step-by-step explanation:

Given : The sum of the squares of 2 consecutive negative integers is 41. What are the numbers.

To find : Which of the following equations is the result of using the factoring method to solve the problem.

Solution : We have given  statement

Let two consecutive number are : n and  n +1 .

Square of two consecutive number are : n² and (n+1)².

According to question :  sum of the squares of 2 consecutive negative integers is 41.

n² + (n+1)² = 41.

n² + n² + 1 +2n =41

2n² + 2n +1 =41

On subtracting 41 from both sides

2n² + 2n +1- 41 = 0

2n² + 2n  - 40 = 0

On dividing by 2 to whole equation

n² + n - 20  = 0

On factoring

n² + 5n -4n - 20  = 0

Taking common n from first two terms and -4 from first two last terms

n (n +5) -4 (n +5) = 0

Grouping

(n -4) (n +5) = 0.

Therefore, Factors are (n + 5)(n - 4) = 0 .

i believe that the answer is

b. y-1=-2(x-4)

hope this !