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+2x−3
Start Fraction x plus 2 over x minus 3 End Fraction

(x+4)(x+2)(x+3)(x−3)
Start Fraction left parenthesis x plus 4 right parenthesis left parenthesis x plus 2 right parenthesis over left parenthesis x plus 3 right parenthesis left parenthesis x minus 3 right parenthesis End Fraction

−(x+2)x+3
Start Fraction negative left parenthesis x plus 2 right parenthesis over x plus 3 End Fraction

−23

Answer 1

To solve the problem, we need to multiply the given rational expressions:

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

Let's start by factoring each of the polynomials:

1. **Factor \(x^2 + 2x - 3\)**:
\[
x^2 + 2x - 3 = (x + 3)(x - 1)
\]

2. **Factor \(x^2 + 3x - 4\)**:
\[
x^2 + 3x - 4 = (x + 4)(x - 1)
\]

3. **Factor \(x^2 + 6x + 8\)**:
\[
x^2 + 6x + 8 = (x + 2)(x + 4)
\]

4. **Factor \(x^2 - 9\)**:
\[
x^2 - 9 = (x - 3)(x + 3)
\]

Now, substitute the factored forms into the expression:

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

Next, we can simplify the expression. Notice that \(x + 3\), \(x - 1\), and \(x + 4\) are present in both the numerator and denominator:

1. The \(x + 3\) in the numerator and denominator cancels.
2. The \(x - 1\) in the numerator and denominator cancels.
3. The \(x + 4\) in the numerator and denominator cancels.

After canceling, we're left with:

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

Thus, the final simplified rational expression is:

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

So, the answer in simplest form is:

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

Alternatively, written as:

Start Fraction x plus 2 over x minus 3 End Fraction.