How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are 5, 2+3i?

1 Answer
Aug 5, 2018

Answer:

#p(x)=x^3-9x^2+33x-65#

Explanation:

#"by the factor theorem"#

#"if "x=a" is a zero then "#

#(x-a)" is a factor of the polynomial"#

#"note that complex zeros occur in conjugate pairs"#

#x=2+3i" is a zero then "2color(red)(-)3i" is also a zero"#

#p(x)=(x-5)(x-(2+3i))(x-(2-3i))#

#color(white)(p(x))=(x-5)(x-2-3i)(x-2+3i)#

#color(white)(p(x))=(x-5)((x-2)^2-(3i)^2)#

#color(white)(p(x))=(x-5)(x^2-4x+4+9)#

#color(white)(p(x))=(x-5)(x^2-4x+13)#

#color(white)(p(x))=x^3-9x^2+33x-65#