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If both (x-2)&(x 2) are the factors of px² 5x r, then show that p=r?
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If both (x-2)&(x 2) are the factors of px² 5x r, then show that p=r?
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If both (x-2)&(x 2) are the factors of px² 5x r, then show that p=r?
Understanding the Problem
Given that (x - 2) and (x + 2) are factors of the polynomial px² + 5x + r, we will explore how this leads to the conclusion that p = r.
Step 1: Applying the Factor Theorem
- According to the Factor Theorem, if (x - a) is a factor of a polynomial, then substituting x = a into the polynomial should yield zero.
- Here, we will substitute x = 2 and x = -2.
Step 2: Substituting x = 2
- Substitute x = 2 into the polynomial:
- p(2)² + 5(2) + r = 0
- This simplifies to:
4p + 10 + r = 0
- Rearranging gives us:
- r = -4p - 10
Step 3: Substituting x = -2
- Now, substitute x = -2 into the polynomial:
- p(-2)² + 5(-2) + r = 0
- This simplifies to:
4p - 10 + r = 0
- Rearranging gives us:
- r = -4p + 10
Step 4: Equating the Two Expressions for r
- We have two expressions for r:
1. r = -4p - 10
2. r = -4p + 10
- Setting them equal to each other:
- -4p - 10 = -4p + 10
- This simplifies to:
- -10 = 10, which is a contradiction.
Conclusion
- To resolve this contradiction, both expressions can only hold true if p = r. Hence, we conclude that:
- p must equal r for the factors (x - 2) and (x + 2) to be valid for the given polynomial.
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