find all zeroes of the polynomial x⁴-3x³-5x²+21x-14 if two zeroes are root 7 and root -7
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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
![](https://hi-static.z-dn.net/files/dee/edbea81f33e60ea486576890286506e4.jpg)
given zeroes is 7 -7
(x-7),(x+7)divide from x⁴-3x³-5x²+21x-14