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Question
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± `sqrt3` , find other zeroes
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Solution 1
Given that 2 + `sqrt3` and 2 - `sqrt3` are zeroes of the given polynomial
Therefore, `(x - 2-sqrt3)(x-2+sqrt3)` = x2 + 4 - 4x - 3
= x2 - 4x + 1 is a factor of the given polynomial
For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing by x4 - 6x3 - 26x2 + 138x - 35 by x2 - 4x + 1
Clearly x4 - 6x3 - 26x2 + 138x - 35 = (x2 - 4x +1)(x2 -2x -35)
It can be observed that (x2 - 2x - 35) is also a factor of the given polynomial.
And (x2 - 2x - 35) = (x -7)(x + 5)
Therefore, the value of the polynomial is also zero when x - 7 = 0 or x + 5 = 0
or x = 7 or -5
Hence, 7 and - 5 are also zeroes of this polynomial
Solution 2
2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35.
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x2 - 4x + 4 = 3,
= (x2 - 4x + 1) (x2 - 2x - 35)
= (x2 - 4x + 1) (x2 - 7x + 5x - 35)
= (x2 - 4x + 1) [x(x - 7) + 5 (x - 7)]
= (x2 - 4x + 1) (x + 5) (x - 7)
∴ (x + 5) and (x - 7) are other factors of p(x).
∴ - 5 and 7 are other zeroes of the given polynomial.
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