Question

Multiply the rational expression and write the product in simplest form: x2+2x−3x2+3x−4⋅x2+6x+8x2−9(1 point)
Responses

(x+4)(x+2)(x+3)(x−3)the fraction with numerator open paren x plus 4 close paren times open paren x plus 2 close paren and denominator open paren x plus 3 close paren times open paren x minus 3 close paren

x+2x−3the fraction with numerator x plus 2 and denominator x minus 3

−(x+2)x+3the fraction with numerator negative open paren x plus 2 close paren and denominator x plus 3

−23

Answer 1

To multiply the rational expressions given, let's start by factoring each quadratic expression separately.

The expressions to consider are:
1. \( x^2 + 2x - 3 \)
2. \( x^2 + 3x - 4 \)
3. \( x^2 + 6x + 8 \)
4. \( x^2 - 9 \)

Now, let's factor each:

1. For \( x^2 + 2x - 3 \):
- The factors are \( (x + 3)(x - 1) \).

2. For \( x^2 + 3x - 4 \):
- The factors are \( (x + 4)(x - 1) \).

3. For \( x^2 + 6x + 8 \):
- The factors are \( (x + 2)(x + 4) \).

4. For \( x^2 - 9 \) (Difference of squares):
- The factors are \( (x + 3)(x - 3) \).

Now, multiplying the rational expressions:

\[
\frac{(x^2 + 2x - 3)(x^2 + 6x + 8)}{(x^2 + 3x - 4)(x^2 - 9)}
\]

becomes:

\[
\frac{(x + 3)(x - 1)(x + 2)(x + 4)}{(x + 4)(x - 1)(x + 3)(x - 3)}
\]

Now we can cancel out the common factors in the numerator and denominator:

- \( (x + 3) \)
- \( (x - 1) \)
- \( (x + 4) \)

After cancellation, we are left with:

\[
\frac{(x + 2)}{(x - 3)}
\]

This fraction is already in its simplest form.

Thus, the final answer is:

\(\frac{x + 2}{x - 3}\)

So the correct response would be:

- \( \frac{x + 2}{x - 3} \) the fraction with numerator \( x + 2 \) and denominator \( x - 3 \)