Questions: Select all the correct answers. Consider this product. (x^2-4x-21)/(3x^2+6x) * (x^2+8x)/(x^2+11x+24) Which values are excluded values for the product? 0 2 -3 -8 7

Solution Steps

The given product is: $$\frac{x^{2}-4 x-21}{3 x^{2}+6 x} \cdot \frac{x^{2}+8 x}{x^{2}+11 x+24}$$ The denominators are $3x^{2} + 6x$ and $x^{2} + 11x + 24$.

Factor $3x^{2} + 6x$: $$3x^{2} + 6x = 3x(x + 2)$$ Factor $x^{2} + 11x + 24$: $$x^{2} + 11x + 24 = (x + 3)(x + 8)$$

The excluded values are the values of $x$ that make any denominator zero. Set each factor equal to zero and solve for $x$:

1. $3x(x + 2) = 0$:
   • $3x = 0 \Rightarrow x = 0$
   • $x + 2 = 0 \Rightarrow x = -2$
2. $(x + 3)(x + 8) = 0$:
   • $x + 3 = 0 \Rightarrow x = -3$
   • $x + 8 = 0 \Rightarrow x = -8$

The excluded values are $0$, $-2$, $-3$, and $-8$. From the given options, the excluded values are $0$, $-3$, and $-8$.

Final Answer