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

Question

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.