We have the following polynomial:
\(f(x)=2x^{3}+9x^{2}+7x-6\)
The problem states that one root is -3. Thus, it is true that \((x+3)\) is a factor of the polynomial. Given that this is fulfilled, it is also true that:
\(f(x)=(x+3)Q(x) \therefore Q(x)=\frac{f(x)}{x+3} \\ \\ where \ Q(x) \ has \ a \ degree \ of \ 2\)
We can find Q(x) by applying Ruffini's rule, thus:
\(\ \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ 7 \ \ \ \ \ \ -6 \\ -3 \\ \rule{50mm}{0.1mm} \\ \ {} \ \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ 3 \ \ \ \ \ -2 \ \ \ \ \ \ \ \ 0\)
Therefore:
\(Q(x)=2x^{2}+3x-2\)
The roots of this polynomial can be get as follows:
\(x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \rightarrow x_{12}=\frac{-3\pm\sqrt{3^2-4(2)(-2)}}{2(2)}\\x_{1}=\frac{1}{2};\ x_{2}=-2\)
These are the roots along with \(-3\). Finally, the factored polynomial can be written as follows:
\(f(x)=(x+3)(x+2)(2x-1)\)
