Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
Use the Distributive Property to prove that these algebraic expressions are equivalent. Explain your answer in one sentence for...Asked by Credance (Dark Sonic)
Use the Distributive Property to prove that these algebraic expressions are equivalent. Explain your answer in one sentence for each expression and show your work.
Expression 1: (x−5)(x+3)
Expression 2: 2x2−x2−2x−20+5
Expression 1: (x−5)(x+3)
Expression 2: 2x2−x2−2x−20+5
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To prove that the expressions \((x - 5)(x + 3)\) and \(2x^2 - x^2 - 2x - 20 + 5\) are equivalent using the Distributive Property, we will expand Expression 1 and simplify Expression 2.
### Step 1: Expanding Expression 1
We will use the Distributive Property (also known as the FOIL method for binomials) to expand \((x - 5)(x + 3)\):
\[
(x - 5)(x + 3) = x(x) + x(3) - 5(x) - 5(3)
\]
\[
= x^2 + 3x - 5x - 15
\]
\[
= x^2 - 2x - 15
\]
### Step 2: Simplifying Expression 2
Now let's simplify Expression 2: \(2x^2 - x^2 - 2x - 20 + 5\):
\[
2x^2 - x^2 - 2x - 20 + 5 = (2x^2 - x^2) - 2x + (-20 + 5)
\]
\[
= x^2 - 2x - 15
\]
### Conclusion
Both expressions simplify to \(x^2 - 2x - 15\), proving that \((x - 5)(x + 3) = 2x^2 - x^2 - 2x - 20 + 5\). Thus, we conclude that the two expressions are equivalent since both yield the same polynomial after simplification.
### Step 1: Expanding Expression 1
We will use the Distributive Property (also known as the FOIL method for binomials) to expand \((x - 5)(x + 3)\):
\[
(x - 5)(x + 3) = x(x) + x(3) - 5(x) - 5(3)
\]
\[
= x^2 + 3x - 5x - 15
\]
\[
= x^2 - 2x - 15
\]
### Step 2: Simplifying Expression 2
Now let's simplify Expression 2: \(2x^2 - x^2 - 2x - 20 + 5\):
\[
2x^2 - x^2 - 2x - 20 + 5 = (2x^2 - x^2) - 2x + (-20 + 5)
\]
\[
= x^2 - 2x - 15
\]
### Conclusion
Both expressions simplify to \(x^2 - 2x - 15\), proving that \((x - 5)(x + 3) = 2x^2 - x^2 - 2x - 20 + 5\). Thus, we conclude that the two expressions are equivalent since both yield the same polynomial after simplification.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.