Question

Consider this product. x^2-4x-21/3x^3+6x x^2+8x/x^2+11x+24 Which values are excluded values for the product?

Answer 1

Using the domain, it is found that x = -8 is excluded for the product.

The domain of a function is the set that contains all possible input values.

• In a fraction, the denominator cannot be 0.

In this problem, there are two fractions, and the points outside the domain, that is, the roots of the denominators, are excluded.

For the first fraction, they are:

$3x^2 + 6x = 0$

$3x(x + 2) = 0$

$3x = 0 \rightarrow x = 0$

$x + 2 = 0 \rightarrow x = -2$

Hence, x = -2(not an option) is excluded.

• x = 0 can be factored with the numerator of the second fraction, so not excluded.

For the second fraction, the roots of the denominator are:

$x^2 + 11x + 24 = 0$

Which is a quadratic equation with coefficients $a = 1, b = 11, c = 24$, hence:

$\Delta = b^2 - 4ac = 11^2 - 4(1)(24) = 121 - 96 = 25$

$x_1 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-11 - 5}{2} = -8$

$x_2 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-11 + 5}{2} = -3$

x = -3 can be factored with the numerator of the first fraction, so only x = -8 is excluded.

A similar problem is given at brainly.com/question/13136492