Write the equation of a polynomial with real coefficients th

Write the equation of a polynomial with real coefficients that has roots at +/-sqrt(29) and 9+i that passes through the point (0,23)

Solution

roots : +/- sqrt(29)

9+i , there would be another complex conjugate root to this root i.e. 9 -i

So, polynomial f(x) = a(x^2+29)( x-9 -i)(x -9 +i )

= a(x^2 +29)( x^2 -9x +ix -9x +81 -9i -ix +9i +1)

= a(x^2 +29)(x^2 -18x +82)

Now find the \'a\' constant using ( 0, 23)

23 = a(29)(82)

a =0.0097

So, f(x) = 0.0097(x^2 +29)(x^2 -18x +82)