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Use the Distributive Property to prove that these algebraic expressions are equivalent. Explain your answer in one sentence for...Asked by .+* Lolbit *+.
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: 2x^2 - x^2 - 2x - 20 + 5
Expression 1: (x - 5) (x + 3)
Expression 2: 2x^2 - x^2 - 2x - 20 + 5
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Answered by
GPT-4o mini
To prove that the two expressions are equivalent using the Distributive Property, we will expand Expression 1 and simplify Expression 2.
**Expression 1: (x - 5)(x + 3)**
Using the Distributive Property (also known as the FOIL method for binomials), we can expand:
\[
(x - 5)(x + 3) = x \cdot x + x \cdot 3 - 5 \cdot x - 5 \cdot 3
\]
\[
= x^2 + 3x - 5x - 15
\]
\[
= x^2 - 2x - 15
\]
So, Expression 1 simplifies to:
**Expression 1 Result:** \(x^2 - 2x - 15\)
**Expression 2: 2x^2 - x^2 - 2x - 20 + 5**
Now, simplify Expression 2:
\[
2x^2 - x^2 - 2x - 20 + 5
\]
\[
= (2x^2 - x^2) - 2x + (5 - 20)
\]
\[
= x^2 - 2x - 15
\]
So, Expression 2 simplifies to:
**Expression 2 Result:** \(x^2 - 2x - 15\)
### Conclusion
Both expressions simplify to \(x^2 - 2x - 15\), thus proving that the expressions \((x - 5)(x + 3)\) and \(2x^2 - x^2 - 2x - 20 + 5\) are equivalent.
**Expression 1: (x - 5)(x + 3)**
Using the Distributive Property (also known as the FOIL method for binomials), we can expand:
\[
(x - 5)(x + 3) = x \cdot x + x \cdot 3 - 5 \cdot x - 5 \cdot 3
\]
\[
= x^2 + 3x - 5x - 15
\]
\[
= x^2 - 2x - 15
\]
So, Expression 1 simplifies to:
**Expression 1 Result:** \(x^2 - 2x - 15\)
**Expression 2: 2x^2 - x^2 - 2x - 20 + 5**
Now, simplify Expression 2:
\[
2x^2 - x^2 - 2x - 20 + 5
\]
\[
= (2x^2 - x^2) - 2x + (5 - 20)
\]
\[
= x^2 - 2x - 15
\]
So, Expression 2 simplifies to:
**Expression 2 Result:** \(x^2 - 2x - 15\)
### Conclusion
Both expressions simplify to \(x^2 - 2x - 15\), thus proving that the expressions \((x - 5)(x + 3)\) and \(2x^2 - x^2 - 2x - 20 + 5\) are equivalent.
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