Jesse and Benjamin are asked to factor 100x^2−100. Here are their solutions. Who is correct? Explain.
Jesse Benjamin
100x^2−100 =(10x)^2−10^2 100x^2−100 =100(x^2−1)
=(10x−10)(10x+10) =100(x−1)(x+1)
=10(x−1)(10)(x+1)
=100(x−1)(x+1)
10(x−1)(10)(x+1)
Both Jesse and Benjamin are correct, but they arrived at different factored forms of the expression.
Jesse factored the expression using the difference of squares formula: (a^2 - b^2) = (a - b)(a + b). In this case, a = 10x and b = 10. Therefore, Jesse factored 100x^2 - 100 as (10x - 10)(10x + 10).
Benjamin factored the expression by noticing that it is a difference of squares where a = x and b = 1. Therefore, Benjamin factored 100x^2 - 100 as 100(x - 1)(x + 1).
Both factored forms are correct and equivalent, but they are written in slightly different ways.
To determine who is correct, we need to compare the two factored forms and check if they are equivalent.
Jesse's solution: (10x-10)(10x+10)
Benjamin's solution: 100(x-1)(x+1)
To check if these two expressions are equivalent, we can expand them and see if they simplify to the same expression.
For Jesse's solution:
(10x-10)(10x+10) = 10(x-1)(10)(x+1) = 100(x-1)(x+1)
For Benjamin's solution:
100(x-1)(x+1) = 100(x^2-1)
As we can see, both forms simplify to the same expression: 100(x^2-1).
Therefore, both Jesse and Benjamin are correct in their factorizations.