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Question
Find all the zeroes of polynomial (2x4 – 11x3 + 7x2 + 13x – 7), it being given that two of its zeroes are `(3 + sqrt(2))` and `(3 - sqrt(2))`.
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Solution
The given polynomial is f(x) = 2x4 – 11x3 + 7x2 + 13x – 7.
Since `(3 + sqrt(2))` and `(3 - sqrt(2))` are the zeroes of f(x) it follows that each one of `(x + 3 + sqrt(2))` and `(x + 3 - sqrt(2))` is a factor of f(x).
Consequently,` [(x - ( 3 + sqrt(2))] [(x - (3 - sqrt(2))]`
= `[(x - 3) - sqrt(2)] [(x - 3) + sqrt(2)]`
= `[(x - 3)^2 - 2]`
= x2 – 6x + 7, which is a factor of f(x).
On dividing f(x) by (x2 – 6x + 7), we get:
`x^2 - 6x + 7")"overline(2x^4 - 11x^3 + 7x^2 + 13x - 7)"("2x^2 + x - 1`
2x4 – 12x3 + 14x2
– + –
x3 – 7x2 + 13x – 7
x3 – 6x2 + 7x
– + –
–x2 + 6x – 7
–x2 + 6x – 7
+ – +
x
f(x) = 0
⇒ 2x4 – 11x3 + 7x2 + 13x – 7 = 0
⇒ (x2 – 6x + 7) (2x2 + x – 7) = 0
⇒ `(x + 3 + sqrt(2)) (x + 3 - sqrt(2)) (2x - 1) (x + 1) = 0`
⇒ `x = -3 - sqrt(2)` or `x = -3 + sqrt(2)` or `x = 1/2` or x = –1
Hence, all the zeroes are `(-3 - sqrt(2)), (-3 + sqrt(2)), 1/2` and –1.
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