Question

# If the zeros of the polynomial f(x)=x3−3x2+x+1 are  (a-b), a and (a+b), find a and b.

Solution

## As the the coefficient of highest power x3 is1, if three roots are α,βandγ, we have f(x)=x3−3x2+x+1 =x3−(α+β+γ)x2+(αβ+γβ+γα)x+αβγ Now let us compare coefficients of similar powers on each side.  First comparing the sum of roots from the coefficient of x2, we have α+β+γ=−ba a−b+a+a+b=−(−31) i.e. 3a=3 i.e.a=1. Also, coefficients of x give us the sum of the product of zeros αβ+γβ+γα=ca=1 a(a−b)+a(a+b)+(a+b)(a−b)=1 i.e. a(a−b+a+b)+a2−b2=1 =2a2+a2−b2 =3a2−b2=1 and as  a=1, b2=3×(1)2−1=2 and  b=±√2 Further, products of roots is (a−b)a(a+b)=1. Hence, a=1 and b=±√2 Note that the two values of b give the same set of roots.

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