Question

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. (5 points) 7, -11, and 2 + 8i

Answer 1

This will be a 4th degree polynomial.  Our root of x = 7 in factorization form is (x-7).  Our root of x = -11 in factorization form is (x+11) and the last one is a complex number.  According to the conjugate root theorem, if we have 2+8i, we also HAVE to have 2-8i.  In factorization form that first one is (x-(2+8i)) which simplifies to (x-2-8i).  Its conjugate in factorization form is (x-2+8i).  Now we will FOIL all that out.  Let's start with the (x-2-8i)(x-2+8i).  That multiplies out to $x^2-2x+8ix-2x+4-16i-8ix+16i-64i^2$.  We have to combine like terms here to shorten that a bit.   $x^2-4x+4-64i^2$.  i^2 is equal to -1, and -1(64) = -64.  Now we have  $x^2-4x+4-(-64)$.  That is  $x^2-4x+68$.  Now let's FOIL in another factorization.  $(x^2-4x+68)(x-7)$.  That comes out to  $x^3-11x^2+96x-476$.  One more term to go!  $x^3-11x^2+96x-476)(x+11)$.  That, finally, is $x^4-25x^2+580x-5236$.