Find the zeros of the polynomial x cube - 15x square 71x -105 , give...
As the zeroes are in AP, let the zeroes be a-d, a, a+d.
The sum of zeroes of the given polynomial is -(-15)/1 = 15.
Therefore, a-d + a + a+d=15. Hence, a = 5.
Also, the product of zeroes is -(-105)/1 = 105.
Therefore, 5(5-d)(5+d)=105, or 25 - d^2 = 21.
Then, d^2 = 4. Hence, d=+-2.
Therefore, the zeroes are 3, 5, 7.
Find the zeros of the polynomial x cube - 15x square 71x -105 , give...
Introduction:
In this problem, we are given a polynomial of degree 3 and we have to find its zeros which are in AP.
Method:
To find the zeros of the polynomial which are in AP, we will use the following steps:
1. Assume the zeros of the polynomial in AP to be a-d, a, a+d where d is the common difference.
2. Using the sum of roots formula, a-d+a+a+d = 15/1 which gives a=5.
3. Substituting the value of a in the polynomial, we get x^3 - 15x^2 + 71x - 105 = (x-5)(x^2 - 10x + 21)
4. The quadratic factor can be factored as (x-3)(x-7) which gives the zeros as 5, 3, and 7.
Conclusion:
The zeros of the given polynomial x^3 - 15x^2 + 71x - 105 in AP are 3, 5, and 7.
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