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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 sqrt(3) ⁢and –sqrt(3).

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प्रश्न

Find all the zeroes of (x4 + x3 – 23x2 – 3x + 60), if it is given that two of its zeroes are `sqrt(3)` and `-sqrt(3)`.

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उत्तर

Let f(x) = x4 + x3 – 23x2 – 3x + 60

Since `sqrt(3)` and `-sqrt(3)` are the zeroes of f(x), it follows that each one of `(x - sqrt(3))` and `(x + sqrt(3))` is a factor of f(x).

Consequently, `(x - sqrt(3)) (x + sqrt(3)) = (x^2 - 3)` is a factor of f(x).

On dividing f(x) by (x2 – 3), we get:

`x^2 - 3")"overline(x^4 + x^3 - 23x^2 - 3x + 60)"("x^2 + x - 20`
            x4            – 3x2                     
            –                +                        
           x3 – 20x2 – 3x + 60
           x3             – 3x
           –                 +                       
                 –20x2 + 60
                 –20x2 + 60
                 +         –                        
                             x                       

f(x) = 0

⇒ (x2 + x – 20) (x2 – 3) = 0

⇒ (x2 + 5x – 4x – 20) (x2 – 3)

⇒ [x(x + 5) – 4(x + 5)] (x2 – 3)

⇒ `(x - 4) (x + 5) (x - sqrt(3)) (x + sqrt(3)) = 0`

⇒ x = 4 or x = –5 or x = `sqrt(3)` or x = `-sqrt(3)`

Hence, all the zeroes are `sqrt(3), -sqrt(3), 4` and –5.

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अध्याय 2: Polynomials - EXERCISE 2B [पृष्ठ ६४]

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आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 2 Polynomials
EXERCISE 2B | Q 17. | पृष्ठ ६४

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