How many different numbers stisfy the equation (x-2)^2(x-1)^3(x+1)^4(x+2)^5 = 0
I really need help. Thanks!
each factor could be zero, thus
x-2 = 0 ---> x = 2
x-1 = 0 ---> x = 1
x+1 = 0 ---> x = -1
x+2 = 0 ---> x = -2
Look how Wolfram graphed it. How many times and where does the graph intersect with the x-axis ?
oops, forgot the link to the graph
http://www.wolframalpha.com/input/?i=%28x-2%29%5E2%28x-1%29%5E3%28x%2B1%29%5E4%28x%2B2%29%5E5+%3D+0
To determine how many different numbers satisfy the equation (x-2)^2(x-1)^3(x+1)^4(x+2)^5 = 0, we need to find the number of distinct roots or solutions.
The equation (x-2)^2(x-1)^3(x+1)^4(x+2)^5 = 0 can be rewritten as a product of factors:
(x-2)^2 = 0
(x-1)^3 = 0
(x+1)^4 = 0
(x+2)^5 = 0
Each factor represents a potential solution to the equation. To find the solutions, we set each factor equal to zero and solve for x:
For the first factor, (x-2)^2 = 0:
(x-2)(x-2) = 0
x-2 = 0
x = 2
For the second factor, (x-1)^3 = 0:
(x-1)(x-1)(x-1) = 0
x-1 = 0
x = 1
For the third factor, (x+1)^4 = 0:
(x+1)(x+1)(x+1)(x+1) = 0
x+1 = 0
x = -1
For the fourth factor, (x+2)^5 = 0:
(x+2)(x+2)(x+2)(x+2)(x+2) = 0
x+2 = 0
x = -2
Therefore, there are four different values of x that satisfy the equation: x = 2, x = 1, x = -1, and x = -2.
Hence, the answer is that there are four different numbers that satisfy the equation (x-2)^2(x-1)^3(x+1)^4(x+2)^5 = 0.