Question

Sum of two zeroes of a polynomial of degree 4 is -1 and their product is -2. If other two zeroes are √3 and -√3, then find the polynomial.

Answer 1

Sum of two zeroes of a polynomial of degree 4 is -1 and their product is -2

Let the zeroes be m and n

m+n = -1
=> m = -(1+n)

mn = -2

=> -(1+n)n = -2

=> (1+n)n= 2

=> n + n^2 - 2 = 0

=> n^2 + 2n - n -2 = 0

=> n(n+2) -1(n+2) = 0

=> (n+2)(n-1) = 0

=> n = -2 or 1

m = -(1+1) or -(1-2) = -2 or 1

So the roots are 1 and -2.

Given that other two roots are √3 and -√3

The polynomial is given as:

$(x - 1)(x + 2)(x - \sqrt{3} )(x + \sqrt{3} ) \\ \\ = ( {x}^{2} + x - 2)( {x}^{2} - 3) \\ \\ = {x}^{4} + {x}^{3} - 5 {x}^{2} - 3x + 6$

Answer 2

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Sum of two zeroes of a polynomial of degree 4 is -1 and their product is -2

Let the zeroes be m and n

m+n = -1

=> m = -(1+n)

mn = -2

=> -(1+n)n = -2

=> (1+n)n= 2

=> n + n^2 - 2 = 0

=> n^2 + 2n - n -2 = 0

=> n(n+2) -1(n+2) = 0

=> (n+2)(n-1) = 0

=> n = -2 or 1

m = -(1+1) or -(1-2) = -2 or 1

So the roots are 1 and -2.

Given that other two roots are √3 and -√3

The polynomial is given as:

$(x - 1)(x + 2)(x - \sqrt{3} )(x + \sqrt{3} ) \\ \\ = ( {x}^{2} + x - 2)( {x}^{2} - 3) \\ \\ = {x}^{4} + {x}^{3} - 5 {x}^{2} - 3x + 6$

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