Find a polynomial function with real coefficients that has the given zeros.

Question

Answer 1

5/9, -1, 5 + sqr2(i)

the imaginary zeroes come in conjugate pairs

if 5+isqr2 is a zero, so is 5-isqr2

x = 5 + or - isqr2

x-5 = + or - isqr2

(x-5)^2 = -2

x^2 -10x +25 +2 = 0

x^2 -10x +27

factors are (x-9/5)(x+1)(x^2 -10x +27)

= (1/5)(5x-9)(x+1)(x^2 -10x+27)

= .2(5x^2-4x-9)(x^2 -10x +27)

= .2(5x^4 -54x^3 +166x^2 -18x -243)

f(x) = x^4 - 10.8x^3 + 33.2x^2 -3.6x - 48.6