To find out the excluded values for the product we only need to analyse the denominator, we can completely ignore the numerator product.
The excluded values are the values that make the denominator equal zero, in other words, it's the zeros of the denominator functions.
Considering only the denominator we have
[tex](3x^2+6x)(x^2+11x+24)[/tex]If we do that product we will get a fourth-degree polynomial, way too hard to solve. But there's an easier way to solve it, the zeros of the two quadratics are the zeros of the product as well, we can calculate the zeros separately, this is the same to calculate the excluded values for each quadratic.
Let's solve this one first:
[tex]3x^2+6x=0[/tex]If we factor x we have
[tex]\begin{gathered} 3x^2+6x\Rightarrow x(3x+6)=0 \\ \\ z_1=0 \\ z_2=-2 \end{gathered}[/tex]Then the zeros are 0 and -2.
Now we solve the other quadratic
[tex]x^2+11x+24=0[/tex]We can solve it using the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}{}}{2a}[/tex]Where a = 1, b = 11 and c = 24.
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}{}}{2a} \\ \\ x=\frac{-11\pm\sqrt[]{11^2-4\cdot24}{}}{2} \\ \\ x=\frac{-11\pm\sqrt[]{25}{}}{2} \\ \\ x=\frac{-11\pm5{}}{2} \\ \\ x_1=\frac{-11+5}{2}=-3 \\ \\ x_2=\frac{-11-5}{2}=-8 \end{gathered}[/tex]Therefore the other two excluded values are -3 and -8. Now we have all excluded values, they are:
[tex]\{0,-2,-3,-8\}[/tex]
