Question

Q1) Find zeroes of polynomial x²-3​

Answer 1

Step-by-step explanation:

Recall the identity a² – b² =(a – b)(a + b). Using it,we can write:

Recall the identity a² – b² =(a – b)(a + b). Using it,we can write:x²-3=(x-√3)(x+√3)

Recall the identity a² – b² =(a – b)(a + b). Using it,we can write:x²-3=(x-√3)(x+√3)So the value of x²-3 is zero when x=√3 or x=-√3

Recall the identity a² – b² =(a – b)(a + b). Using it,we can write:x²-3=(x-√3)(x+√3)So the value of x²-3 is zero when x=√3 or x=-√3Therefore, the zeroes of x²-3 are √3 and -√3

Recall the identity a² – b² =(a – b)(a + b). Using it,we can write:x²-3=(x-√3)(x+√3)So the value of x²-3 is zero when x=√3 or x=-√3Therefore, the zeroes of x²-3 are √3 and -√3Sum of zeroes =√3-√3=-3/1=-(Coefficient of x)/(Coefficient of x²)

Recall the identity a² – b² =(a – b)(a + b). Using it,we can write:x²-3=(x-√3)(x+√3)So the value of x²-3 is zero when x=√3 or x=-√3Therefore, the zeroes of x²-3 are √3 and -√3Sum of zeroes =√3-√3=-3/1=-(Coefficient of x)/(Coefficient of x²)Product of zeroes =(√3)(-√3)=(-3)/1 =(Constant term)/(Coefficient of x²

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Answer 2

Step-by-step explanation:

this is a solution of polynomial.

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