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Find All the Zeroes of `(X^4 + X^3 – 23x^2 – 3x + 60)`, If It is Given that Two of Its Zeroes Are `Sqrt3 and –Sqrt3`. - Mathematics

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Find all the zeroes of `(x^4 + x^3 – 23x^2 – 3x + 60)`, if it is given that two of its zeroes are `sqrt3 and –sqrt3`. 

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Solution

Let f(x) =`x^4 + x^3 – 23x^2 – 3x + 60`
Since `sqrt3` and `–sqrt3` are the zeroes of f(x), it follows that each one of (x – √3) and (x + √3) is a factor of f(x).
Consequently, `(x – sqrt3) (x + sqrt3) = (x^2 – 3)`is a factor of f(x).
On dividing f(x) by `(x^2 – 3)`, we get:  

  

f(x) = 0
⇒ `(x^2 + x – 20) (x^2 – 3) = 0`
⇒ `(x^2 + 5x – 4x – 20) (x^2 – 3)`
⇒ `[x(x + 5) – 4(x + 5)] (x^2 – 3)`
⇒ `(x – 4) (x + 5) (x – sqrt3) (x + sqrt3) = 0`
⇒ `x = 4 or x = -5 or x = sqrt3 or x = -sqrt3`
Hence, all the zeroes are √3, -√3, 4 and -5.  

 

 

Concept: Relationship Between Zeroes and Coefficients of a Polynomial
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