x^2-2√5x+3

= x^2-{(√5+√2) + (√5-√2)} x + (√5-√2) (√5+√2)

=x²-(√5 + √2)x - (v5-v2)x+ (√5 + √2)(√5-√2) =x[x - √5-√2)-(v5-√2)(x-v5-v2]

=(x-v5-v2][x - √5 + √2]

Hence, x = √5 ± √2 are the zeros of given expression.

Hope this helped you out❤️

Sorry, I have been A. W. O. L. of late...

Answer:

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