Math, asked by gauriabeena, 29 days ago

find all zeroes of the polynomial x⁴-3x³-5x²+21x-14 if two zeroes are root 7 and root -7​

Answers

Answered by aqimshaik
1

Answer:

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Step-by-step explanation:

your Solution Given here ✔✔

let x43x³ - 5x² +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 + √7) -√√7(x + √7) => x² +√7x - √7x+7 => x² -7 so, x^2 - 7 is a factor of f (x) dividing f (x) by x^2 - 7

see in above attachment. 111

=> f(x)=0

=> (x² ·3x + 2) (x + √7)(x − √7) = - => (x² − 2x − 1x + 2)(x + √7)(x − √i - - => x(x-2)-1(x-2)(x+√7)(x - √^ => (x-2)(x - 1)(x+√7)(x - √7) => 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 -

-√7, √7, 2 and 1

Attachments:
Answered by anudghufcvb
0

given zeroes is 7 -7

(x-7),(x+7)divide from x⁴-3x³-5x²+21x-14

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