## College Algebra (6th Edition)

solution set: $\displaystyle \{-\frac{1}{2},\ 1,\ 2\}$
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 $f(c)$. (synthetic division: the last number in the last row is $f(c)$ ) ------------------- The problem, in fact, asks us to find zeros of f. First, let's divide $(2x^{3}-5x^{2}+x+2)\div(x-2)\qquad (c=2)$ $\begin{array}{lllllll} \underline{2}| & 2 & -5 & 1 & 2 & & \\ & & 4 & -2 & -2 & & \\ & -- & -- & -- & -- & -- & \\ & 2 & -1 & -1 & |\overline{ 0 } & & \end{array}$ So, $(x-2)$ is a factor of $f(x).$ $f(x)=(x-2)(2x^{2}-x-1)$ we try to factor the trinomial by searching for two integers whose product is $2\cdot(-1)=-2$ and whose sum is $-1$.... We find $1$ and $-2:\quad$ $2x^{2}-x-1= 2x^{2}+x-2x-1$ ... factor in pairs ... $=x(2x+1)-(2x+1)$ $=(2x+1)(x-1)$ ... factor out 2 from the first parentheses ... $=2(x+\displaystyle \frac{1}{2})(x-1)$ So, $f(x)=2(x+\displaystyle \frac{1}{2})(x-1)(x-2)$ and by (b) of the Factor Theorem, the solution set of the equation $f(x)=0$ are the zeros of f: $\displaystyle \{-\frac{1}{2},\ 1,\ 2\}$