MHB Find Zeros of $2x^3+5x^2-28x-15$ with Synthetic Division

[karush]

$\text{factor}$
$g(x)=2x^3+5x^2-28x-15$
$\text{synthetic division}$
$\begin{array}{c|rr}
& 2 & 5 &-28 &-15 \\
&&6&33 & 15\\ \hline
3&2&11&5&0
\end{array}$
$\text{thus }\\$
$2x^2+11x+5$
$\text{use quadradic formula}$
$\begin{align*}
x=\frac{-(11)\pm\sqrt{(11)^2-4(2)(5)}}{2(1)}
&=\frac{-11\pm\sqrt{81}}{2}
=\frac{-11\pm9}{2}
\end{align*}$
$x=-10,-1 \quad g(x)=(x+10)(x+1)(x-3)$
hopefully
comment?

 

[MarkFL]

Check your application of the quadratic formula, or simply factor. :)

 

[karush]

a=2 not 1

use quadradic formula $\begin{align*} x=\frac{-(11)\pm\sqrt{(11)^2-4(2)(5)}}{2(2)} &=\frac{-11\pm\sqrt{81}}{4} =\frac{-11\pm9}{4} \end{align*}$ $\text{then $x=-5,-\frac{1}{2}$ so $g(x)=(x+10)(x+\frac{1}{2})(x-3)$}$