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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)`. - Mathematics

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प्रश्न

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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उत्तर

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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अध्याय 2: Polynomials - Exercises 2

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आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 2 Polynomials
Exercises 2 | Q 19

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