Answer
solution set: $\displaystyle \{-\frac{1}{2},\ 1,\ 2\} $
Work Step by Step
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 $f(c)$ )
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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\} $
