Respuesta :
We know that the polynomial have zeros at 3 and 2i; now since it has an imaginary zero this means that its conjugate is also a zero, that is, -2i has to be a zero as well. We know that a polynomial function with zeros a, b and c can be express as:
[tex]f(x)=A(x-a)(x-b)(x-c)[/tex]where A is some constant determine by some other conditions.
In this case the polynomial will have the form:
[tex]f(x)=A(x-2i)(x+2i)(x-3)=A(x^2+4)(x-3)[/tex]Now, to determine the value of A we use the fact that the polynomial passes through the point (4,10); this means that when x=4 the function has to be equal to 10, then we have:
[tex]\begin{gathered} A(4^2+4)(4-3)=10 \\ A(20)(1)=10 \\ 20A=10 \\ A=\frac{10}{20} \\ A=\frac{1}{2} \end{gathered}[/tex]Therefore the function we are looking for is:
[tex]f(x)=\frac{1}{2}(x^2+4)(x-3)[/tex]To find the leading coefficient we expand the expression, then we have:
[tex]\begin{gathered} f(x)=\frac{1}{2}(x^3-3x^2+4x-12) \\ =\frac{1}{2}x^3-\frac{3}{2}x^2+2x-6 \end{gathered}[/tex]Therefore the leading coefficient of the function is equal to 1/2
