Question

find all zeroes of polynomial x4-3x3-5x2+21x-14,if two of it's zeroes are root 7 and root -7.

Answer 1

Hello Dear friend.

your Solution Given here ✔✔
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$let \: \: \: {x}^{4} - 3 {x}^{3} - 5 {x}^{2} + 21x - 14$
since , √7 and -√7 are the zeroes of f(x)
it follows that (x - √7) and (x + √7) are the factor of f(x)

Consequently = ( x - √7 ) and ( x + √7)
$= > x(x + \sqrt{7} ) - \sqrt{7} (x + \sqrt{7} ) \\ = > {x}^{2} + \sqrt{7} x - \sqrt{7} x + 7 \\ = > {x}^{2} - 7$
so, x^2 - 7 is a factor of f (x)
dividing f (x) by x^2 - 7

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see in above attachment. ⬆⬆⬆
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=> f ( x )=0
$= > ( {x}^{2} - 3x + 2)(x + \sqrt{7} )(x - \sqrt{7} ) = 0 \\ = > ({x}^{2} - 2x - 1x + 2)(x + \sqrt{7} )(x - \sqrt{7} ) = 0 \\ = > x(x - 2) - 1(x - 2)(x + \sqrt{7} )(x - \sqrt{7}) = 0 \\ = > (x - 2)(x - 1)(x + \sqrt{7} )(x - \sqrt{7} ) = 0$
=> x - 2 =0 , x - 1 =0 , x + √7 =0 or x - √7 = 0
=> x = 2 , x =1 , x = -√7 or x = √7

Hence , the zeroes of given Polynomial are ―

$- \sqrt{7}, \: \sqrt{7} ,\: \: 2 \: \: and \: 1$
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Hope it's helps you
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Answer 2

Answer:

since , √7 and -√7 are the zeroes of f(x)

it follows that (x - √7) and (x + √7) are the factor of f(x)

Consequently = ( x - √7 ) and ( x + √7)

so, x^2 - 7 is a factor of f (x)

dividing f (x) by x^2 - 7

=> f ( x )=0

= > ( {x}^{2} - 3x + 2)(x + \sqrt{7} )(x - \sqrt{7} ) = 0 \\ = > ({x}^{2} - 2x - 1x + 2)(x + \sqrt{7} )(x - \sqrt{7} ) = 0 \\ = > x(x - 2) - 1(x - 2)(x + \sqrt{7} )(x - \sqrt{7}) = 0 \\ = > (x - 2)(x - 1)(x + \sqrt{7} )(x - \sqrt{7} ) = 0 \\  

=> x - 2 =0 , x - 1 =0 , x + √7 =0 or x - √7 = 0

=> x = 2 , x =1 , x = -√7 or x = √7

Hence , the zeroes of given Polynomial are ―

- \sqrt{7}, \: \sqrt{7} ,\: \: 2 \: \: and \: 1

Step-by-step explanation: