Write the rquation in standard form of a polynomial with rea

Write the rquation, in standard form, of a polynomial with realcoefficients that has roots at -3, -4 + 3i, and 2 + 5i, and passes through the point (0, -33). You answer shold include only (decimal) numbers, the letter \"x\" and th characters \"+\", \"-\", and \"^\".

f(x) =

Solution

Let the equation of the function be y = f (x) = a (x +3) (x + 4 -3i)( x – 2 -5i). Since the graph of the polynomial passes through ( 0, -33), we have -33 = a* 3 * (4-3i)( -2 -5i) or, a = -33/ [3 * (4-3i)( -2 -5i) ] or, a = -11(4 +3i)(-2 +5i)/ [(4-3i)( 4 + 3i)(-2 -5i)(-2 + 5i) = -11(4 +3i)(-2 +5i)/ ( 16 -9i²) ( 4 – 25i²) =                       (-23+ 14i)/( 25*29) = -23/725 + 14i/725 = - 0.0317 + 0.0193i ( rounded off to 4 decimal places.

Thus, the equation of the polynomial is y = f(x) = (- 0.0317 + 0.0193i) ( x +3) (x + 4 -3i) ( x – 2 -5i) or,

y = (- 0.0317 + 0.0193i)( x³ + 5x² – 8 ix² -17x -38 ix -69 -42i) = - 0.0317x³- 0.1585x² + 0.02536 ix² + 0.5380x + 1.2046ix + 2.1873 + 1.3314i + 0.0193i x³ + 0.0965 i x² + 0.1544x² – 0.3281ix + 0.7334x -1.3317i + 0.8106

or, y = f (x ) - 0.0317x³ + 0.0193i x³ - 0.0041x²+0.12186ix² + 1.2714x + 0.8765ix + 2.9979 – 0.0003 i

It is not possible for the function to have only real coefficients.