How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are 4, 4, 2+i?

Question

Chloe Alexander · Accepted Answer

#f(x) = x^4-12x^3+53x^2-104x+80# Assuming you want a polynomial with real coefficients, the complex conjugate #2-i# must also be a zero and the monic polynomial of least degree is: #f(x) = (x-4)(x-4)(x-(2+i))(x-(2-i))# #color(white)(f(x)) = (x^2-8x+16)((x-2)+i)((x-2)-i)# #color(white)(f(x)) = (x^2-8x+16)((x-2)^2-i^2)# #color(white)(f(x)) = (x^2-8x+16)(x^2-4x+4+1)# #color(white)(f(x)) = (x^2-8x+16)(x^2-4x+5)# #color(white)(f(x)) = x^4-12x^3+53x^2-104x+80# Any polynomial in #x# with these zeros will be a multiple (scalar or polynomial) of this #f(x)#.

Xavier Alonso · Answer

undefined To write a polynomial function with the given zeros (4, 4, 2+i), you can use the fact that complex roots always come in conjugate pairs. Therefore, if 2+i is a zero, its conjugate 2-i must also be a zero.

The polynomial function with the given zeros can be written as:

$f(x) = (x - 4)(x - 4)(x - (2 + i))(x - (2 - i))$

Simplify this expression to get the polynomial function in its factored form. Then, if you want to find the polynomial function in standard form, you'll need to multiply out the factors.