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find the zeroes of the polynomial f of x = x cube -12×Sq +39× - 28 ,if it is given that the zeroes are in AP.
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find the zeroes of the polynomial f of x = x cube -12×Sq +39× ...
Finding the Zeroes of the Polynomial
To find the zeroes of the polynomial f(x) = x³ - 12x² + 39x - 28, given that the zeroes are in Arithmetic Progression (AP), follow these steps:
Step 1: Understanding Zeroes in AP
- Let the zeroes be a - d, a, and a + d.
- The sum of the zeroes can be represented as: (a - d) + a + (a + d) = 3a.
Step 2: Applying Vieta's Formulas
- According to Vieta's formulas, the sum of the zeroes is equal to the coefficient of x² (with opposite sign):
- 3a = 12, hence a = 4.
Step 3: Finding the Zeroes
- The zeroes can now be expressed as:
- 4 - d, 4, and 4 + d.
- The product of the zeroes (from Vieta's) is given by:
- (4 - d) * 4 * (4 + d) = 28.
Step 4: Simplifying the Product
- Expanding the equation:
- (16 - d²) * 4 = 28
- 64 - 4d² = 28
- 4d² = 36
- d² = 9, thus d = 3 or d = -3.
Step 5: Final Zeroes
- Therefore, the zeroes are:
- 4 - 3 = 1,
- 4,
- 4 + 3 = 7.
Conclusion
- The zeroes of the polynomial f(x) = x³ - 12x² + 39x - 28 are:
- 1, 4, 7, which are indeed in AP.
To find the zeroes of the polynomial f(x) = x³ - 12x² + 39x - 28, given that the zeroes are in Arithmetic Progression (AP), follow these steps:
Step 1: Understanding Zeroes in AP
- Let the zeroes be a - d, a, and a + d.
- The sum of the zeroes can be represented as: (a - d) + a + (a + d) = 3a.
Step 2: Applying Vieta's Formulas
- According to Vieta's formulas, the sum of the zeroes is equal to the coefficient of x² (with opposite sign):
- 3a = 12, hence a = 4.
Step 3: Finding the Zeroes
- The zeroes can now be expressed as:
- 4 - d, 4, and 4 + d.
- The product of the zeroes (from Vieta's) is given by:
- (4 - d) * 4 * (4 + d) = 28.
Step 4: Simplifying the Product
- Expanding the equation:
- (16 - d²) * 4 = 28
- 64 - 4d² = 28
- 4d² = 36
- d² = 9, thus d = 3 or d = -3.
Step 5: Final Zeroes
- Therefore, the zeroes are:
- 4 - 3 = 1,
- 4,
- 4 + 3 = 7.
Conclusion
- The zeroes of the polynomial f(x) = x³ - 12x² + 39x - 28 are:
- 1, 4, 7, which are indeed in AP.
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find the zeroes of the polynomial f of x = x cube -12×Sq +39× - 28 ,if it is given that the zeroes are in AP. Related: Video: Introduction to Polynomials (Detailed)?
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