## College Algebra (10th Edition)

All zeros of h: $3-2i\quad$(given), $3+2i,\ -2$, and $5.$
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 $u=3-2i$ is a zero, its conjugate $v=3+2i$ is also a zero. $b=-(u+v)=-(3-2i+3+2i)=-6$ $c=(3-2i)(3+2i)=9+4=13$ A quadratic binomial with these zeros is $x^{2}-6x+13 ,$ and, having the same factors, it is a factor of h. Dividing $(x^{4}-9x^{3}+21x^{2}+21x-130)\div(x^{2}-6x+13)$ $\left[\begin{array}{l} \\ x^{2}-6x+13\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{array}\right. \left.\begin{array}{lllll} x^{2} & -3x & -10 & & \\ \hline)\ x^{4} & -9x^{3} & +21x^{2} & +21x-130 & \\ x^{2} & -6x^{3} & +13x^{2} & & \\ -- & -- & -- & & \\ & -3x^{3} & +8x^{2} & +21x-130 & \\ & -3x^{3} & +18x^{2} & -39x & \\ & -- & -- & -- & \\ & & -10x^{2} & +60x-130 & \\ & & -10x^{2} & +60x-130 & \\ & & -- & -- & \\ & & & 0 & \end{array}\right.$ $h(x)=(x^{2}-6x+13)(x^{2}-3x-10)$ ... two factors of -10 whose sum is -3 are -5 and +2 ... $h(x)=(x^{2}-6x+13)(x-5)(x+2)$ All zeros of h: $3-2i\quad$(given), $3+2i,\ -2$, and $5.$