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Question
Find all the zeroes of polynomial `(2x^4 – 11x^3 + 7x^2 + 13x – 7)`, it being given that two of its zeroes are `(3 + sqrt2) and (3 – sqrt2)`.
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Solution
The given polynomial is f(x) = `2x^4 – 11x^3 + 7x^2 + 13x – 7.`
Since `(3 + sqrt2) and (3 – sqrt2)` are the zeroes of f(x) it follows that each one of `(x + 3 + sqrt2) and (x + 3 – sqrt2) `is a factor of f(x).
Consequently,` [(x – ( 3 + sqrt2)] [(x – (3 – sqrt2)] = [(x – 3) - sqrt2 ] [(x – 3) + sqrt2 ]`
=`[(x – 3)^2 – 2 ] = x^2 – 6x + 7,` which is a factor of f(x).
On dividing f(x) by `(x^2 – 6x + 7)`, we get:
f(x) = 0
⇒` 2x^4 – 11x^3 + 7x^2 + 13x – 7 = 0`
⇒ `(x^2 – 6x + 7) (2x2 + x – 7) = 0`
⇒` (x + 3 + sqrt2) (x + 3 – sqrt2) (2x – 1) (x + 1) = 0`
⇒ `x = –3 – sqrt2 or x = –3 + sqrt2 or x =1/2 or x = -1`
Hence, all the zeroes are `(–3 – sqrt2), (–3 + sqrt2),1/2 and -1.`
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