Question

Given that 3-4i is a zero, factor the following polynomial function completely. f(x)=x^4-17x^3+115x^2-419x+600

Answer 1

Since the complex number 3 - 4i is a zero, so its conjugate, 3 + 4i, is also a zero.

Then, to factor this polynomial function, let's find the other 2 zeros of the function.

To do so, let's divide the polynomial by another polynomial created by the two known roots:

`\begin{gathered} (x-3+4i)(x-3-4i) \\ =x^2-3x-4ix-3x+9+12i+4ix-12i+16 \\ =x^2-6x+25 \end{gathered}`

Now, dividing the polynomials, we have:

x^4 divided by x^2: x^2

x^2 times (x^2 - 6x + 25): x^4 - 6x^3 + 25x^2

x^4-17x^3+115x^2-419x+600 minus x^4 - 6x^3 + 25x^2: -11x^3 + 90x^2 - 419x + 600

-11x^3 divided by x^2: -11x

-11x times (x^2 - 6x + 25): -11x^3 + 66x^2 - 275x

-11x^3 + 90x^2 - 419x + 600 minus -11x^3 + 66x^2 - 275x: 24x^2 - 144x + 600

24x^2 divided by x^2: 24

24 times (x^2 - 6x + 25): 24x^2 - 144x + 600

24x^2 - 144x + 600 minus 24x^2 - 144x + 600: 0

So the result of the division is x^2 - 11x + 24.

Finding its zeros, we have:

`\begin{gathered} a=1,b=-11,c=24 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_1=\frac{11+\sqrt[]{121-96}}{2}=(11+5)/(2)=8 \\ x_2=(11-5)/(2)=3 \end{gathered}`

Finally, putting f(x) in the factored form, we have:

`\begin{gathered} f(x)=a(x-x_1)(x-x_2)(x-x_3)(x-x_4) \\ f(x)=(x-3+4i)(x-3-4i)(x-3)(x-8) \end{gathered}`