## College Algebra (6th Edition)

zeros of f: $-1,\ 2,\ 3$
We will use The Factor Theorem Let $f(x)$ be a polynomial. $\mathrm{a}$. If $f(c)=0$, then $x-c$ is a factor of $f(x)$. $\mathrm{b}$. If $x-c$ is a factor of $f(x)$, then $f(c)=0\quad$(c is a zero of f). Also, The Remainder Theorem If the polynomial $f(x)$ is divided by $x-c$, then the remainder is $f(c)$. (in synthetic division, the last number in the last row is $f(c)$ ) ------------------- First, let's divide $(x^{3}-4x^{2}+x+6)\div(x+1)\qquad (c=-1)$ $\begin{array}{lllllll} \underline{-1}| & 1 & -4 & 1 & 6 & & \\ & & -1 & 5 & -6 & & \\ & -- & -- & -- & -- & -- & \\ & 1 & -5 & 6 & |\overline{ 0 } & & \end{array}$ So, $(x+1)$ is a factor of $f(x).$ $f(x)=(x+1)(x^{2}-5x+6)$ we try to factor the trinomial by searching for two integers whose product is $6$ and whose sum is $-5$.... We find $-2$ and $-3:\quad x^{2}-5x+6=(x-2)(x-3)$ So, f(x) has three factors, $f(x)=(x+1)(x-2)(x-3)$, and by (b) of the Factor Theorem, zeros of f: $-1,\ 2,\ 3$