## College Algebra (11th Edition)

$f(2+i) =13+7i \; \neq 0$, so $k=2+i$ is not a zero of $f(x).$
The remainder theorem$:$ If we divide a polynomial $f(x)$ with $(x-k)$, the remainder is $f(k)$. If k is a zero, then $f(k)=0$ (the remainder of the above division is 0). Set up the synthetic division table: When dividing with ($x-\mathrm{k})$, place $k$ to the left of the bar in row 1 List all coefficients of the dividend (the numerator), including 0's, right of the bar in row 1. Row two is empty for now, we fill it step by step. Bottom row: copy the leading coefficient $\left\{\begin{array}{lllllll} k & | & a_{n} & a_{n-1} & ... & a_{1} & a_{0}\\ & & & \fbox{$A=?$} & & & \\ & & -- & -- & -- & -- & --\\ & & a_{n} & \fbox{$B=?$} & & & \end{array}\right.$ In each step, calculate A and B so that $A=k\times$(last calculated coefficient in bottom row), $B=$ sum of two numbers above. --- $k=2+i$ First step: $\left\{\begin{array}{lllll} 2+i & | & 1 & 3 & 4\\ & & & \fbox{$A=?$} & \\ & & -- & -- & --\\ & & 1 & \fbox{$B=?$} & \end{array}\right.$ Second step: $\left\{\begin{array}{lllll} 2+i & | & 1 & 3 & 4\\ & & & 2+i & \fbox{$A=?$}\\ & & -- & -- & --\\ & & 1 & 5+i & \fbox{$B=?$} \end{array}\right.$ ... repeat this process... $\left\{\begin{array}{lllll} 2+i & | & 1 & 3 & 4\\ & & & 2+i & 9+7i\\ & & -- & -- & --\\ & & 1 & 5+i & \fbox{$13+7i$} \end{array}\right.\qquad\left(\begin{array}{l} (2+i)(5+i)=\\ =10+7i+i^{2}\\ =10+7i-1\\ =9+7i \end{array}\right)$ Interpret bottom row: The last number is the remainder. The rest are coefficients of the quotient, whose degree is 1 less than the dividend. $f(2+i) =13+7i \; \neq 0$, so $k=2+i$ is not a zero of $f(x).$