## College Algebra (10th Edition)

$P(x)=x^4-14x^3+77x^2-200x+208$
RECALL: If $a+bi$ is a zero of a polynomial function with real number coefficients, then its conjugate $a-bi$ is also a zero of the function. Thus, the missing zero of the given function is the conjugate of $3+2i$, which is $\color{blue}{\bf3-2i}$. The zeros of the polynomial function with real coefficients are: $3+2i \\3-2i \\4 \\4$ This means that the function is: $P(x) = a[x-(3+2i)] \cdot [x-(3-2i)] \cdot (x-4)(x-4) \\P(x) = a(x-3-2i)(x-3+2i)(x-4)^2 \\P(x) =a[(x-3)-2i]\cdot[(x-3)+2i] (x^2-8x+16) \\P(x) = a[(x-3)^2-4i^2](x^2-8x+16) \\P(x) = a[x^2-6x+9-4(-1)](x^2-8x+16) \\P(x) = a(x^2-6x+9+4)(x^2-8x+16) \\P(x)=a(x^2-6x+13)(x^2-8x+16) \\P(x) = a(x^4-8x^3+16x^2-6x^3+48x^2-96x+13x^2-104x+208) \\P(x)=a(x^4-14x^3+77x^2-200x+208)$ Setting the leading coefficient $a=1$ gives: $P(x)=x^4-14x^3+77x^2-200x+208$