5.12 synthetic division

[karush]

Use factor theorem and syntheitc division and its conjugate to decide whether the second polynomial is a factor of the first
$x^3-5x^2+3x+1;\quad x+1$
\item \textit{apply synthetic division}
\item$\begin{array}{c|rrrrr}
1 &1 &-5 &3 &1\\
& &1 &-4 &-1\\
\hline &1 &-4 &-1 &0
\end{array}$
$(x-1)$ so $x^2-4x-1$
$\begin{array}{rl}
x &=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\
\textsf{a,b,c} &=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(-1)}}{2(-1)}
=\dfrac{4\pm\sqrt{20}}{-2}
=\dfrac{4\pm 2\sqrt{5}}{2}
=2+\sqrt{5}\\
\textsf{hence} &x=1,-1,2+\sqrt{5}
\end{array}$

my first pass thru this...
actually I didn't get what the conjugate thing was about?

 

[skeeter]

zeros of $f(x) = x^3 -5x^2 + 3x + 1$ are $x \in \{ 2- \sqrt{5}, 1 , 2+\sqrt{5} \}$

-1]  1  -5  3   1
        -1  6  -9
------------------
     1  -6  9  -8

$f(-1) = -8 \implies (x+1)$ is not a factor of the cubic polynomial

 

[HOI]

If I wanted to determine whether or not x+ 1 is a factor of $x^3- 5x^2+ 3x+ 1$ I would simply observe that when x= -1, $(-1)^3- 5(-1)^2+ 3(-1)+ 1= -1- 5- 3+ 1= -9+ 1= -8$. Since that is not 0, no, x+1 is NOT a factor of $x^3- 5x^2+ 3x+ 1$.