# Conjugate pairsBuild a polynomial whose roots are 1 and 5i.

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A polynomial that has complex roots always has them in sets of complex conjugate numbers. As the given polynomial has 1 and 5i as roots, we know that -5i is also a root. The polynomial is: ( x - 1)(x - 5i)(x + 5i)

=> (x^2 + 25)(x - 1)

=> x^3 - x^2 + 25x - 25

Since 5i is a root, then its conjugate, −5i is also the root of the polynomial.

They represent the roots of a quadratic factor of the polynomial.

Now, we'll apply Viete's relations to determine the quadratic factor.

The sum of those roots is S = 5i-5i = 0. The product is P = 5i*(-5i) = -25i^2 = 25 (i^2 = -1).

Therefore the quadratic factor is:

x^2 + 25

Next, since 1 is a root, then (*x* − 1) is a factor. Therefore the polynomial is:

(x − 1)(x^2 + 25) = x^3 −x^2 + 25x − 25.