## College Algebra (10th Edition)

All zeros of f: $1+3i\quad$(given), $1-3i,\ -1$, and $6.$
If $u$ and $v$ are zeros of $x^{2}+bx+c ,$ then $(x-u)(x-v)=x^{2}-(u+v)x+uv$ equals $x^{2}+bx+c ,$ from where we extract: $b=-(u+v),\quad c=uv$ Since $u=1+3i$ is a zero, its conjugate $v=1-3i$ is also a zero. $b=-(u+v)=-(1+3i+1-3i)=-2$ $c=(1+3i)(1-3i)=1+9=10$ A quadratic binomial with these zeros is $x^{2}-2x+10 ,$ and, having the same factors, it is a factor of f. Dividing $(x^{4}-7x^{3}+14x^{2}-38x-60)\div(x^{2}-2x+10)$ ${\small \left[\begin{array}{l} \\ x^{2}-2x+10\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{array}\right. \left.\begin{array}{lllll} x^{2} & -5x & -6 & & \\ \hline)\ x^{4} & -7x^{3} & +14x^{2} & -38x-60 & \\ x^{4} & -2x^{3} & +10x^{2} & & \\ -- & -- & -- & & \\ & -5x^{3} & +4x^{2} & -38x-60 & \\ & -5x^{3} & +10x^{2} & -50x & \\ & -- & -- & -- & \\ & & -6x^{2} & +12x-60 & \\ & & -6x^{2} & +12x-60 & \\ & & -- & -- & \\ & & & 0 & \end{array}\right]}$ $h(x)=(x^{2}-2x+10)(x^{2}-5x-6)$ ... two factors of $-6$ whose sum is $-5$ are $-6$ and $+1$ ... $h(x)=(x^{2}-2x+10)(x-6)(x+1)$ All zeros of f: $1+3i\quad$(given), $1-3i,\ -1$, and $6.$