Question

Hello, May I please get some assistance with this homework question? I posted an image below Q11

Answer 1

The formula of the polynomial f(x) has the form:

`f(x)=a(x-x_1)^n(x-x_2)^m\ldots`

where a is the leading coefficient, x₁, x₂, etc, are the zeros, and n, m, etc are the multiplicity of each zero.

Given that 5+5i is a zero, then its conjugate 5-5i is also a zero. With these zeros and the zero: 3 of multiplicity 2, we get the next expression:

`f(x)=a(x-3)^2(x-(5+5i))(x-(5-5i))`

which has a degree of 4.

Expanding the expression, we get:

`\begin{gathered} f(x)=a\lbrack x^2-6x+9\rbrack\lbrack x^2-x(5-5i)-(5+5i)x+(5+5i)(5-5i)\rbrack \\ f(x)=a\lbrack x^2-6x+9\rbrack\lbrack x^2-5x+5ix-5x-5ix+(5\cdot5-5\cdot5i+5i\cdot5-(5i)^2)\rbrack \\ f(x)=a\lbrack x^2-6x+9\rbrack\lbrack x^2-10x+50\rbrack \\ f(x)=a(x^2\cdot x^2-x^2\cdot10x+x^2\cdot50-6x\cdot x^2+6x\cdot10x-6x\cdot50+9x^2-9\cdot10x+9\cdot50) \\ f(x)=a(x^4-10x^3+50x^2-6x^3+60x^2-300x+9x^2-90x+450) \\ f(x)=a(x^4-16x^3+119x^2-390x+450) \end{gathered}`