Question
Multiply the rational expression and write the product in simplest form:
x^2+2x−3/x^2+3x−4⋅x^2+6x+8/x^2−9
(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+2)/x+3
the fraction with numerator negative open paren x plus 2 close paren and denominator x plus 3
x+2/x−3
the fraction with numerator x plus 2 and denominator x minus 3
−2/3
negative 2 thirds
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x^2+2x−3/x^2+3x−4⋅x^2+6x+8/x^2−9
(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+2)/x+3
the fraction with numerator negative open paren x plus 2 close paren and denominator x plus 3
x+2/x−3
the fraction with numerator x plus 2 and denominator x minus 3
−2/3
negative 2 thirds
Skip to navigation
page 16 of 16
Answers
GPT-4o mini
To solve the expression
\[
\frac{x^2 + 2x - 3}{x^2 + 3x - 4} \cdot \frac{x^2 + 6x + 8}{x^2 - 9},
\]
we will simplify each rational expression before multiplying.
**Step 1: Factor each polynomial.**
1. For \(x^2 + 2x - 3\):
\[
x^2 + 2x - 3 = (x + 3)(x - 1)
\]
2. For \(x^2 + 3x - 4\):
\[
x^2 + 3x - 4 = (x + 4)(x - 1)
\]
3. For \(x^2 + 6x + 8\):
\[
x^2 + 6x + 8 = (x + 2)(x + 4)
\]
4. For \(x^2 - 9\) (difference of squares):
\[
x^2 - 9 = (x + 3)(x - 3)
\]
**Step 2: Substitute the factors back into the expression.**
Now we rewrite the expression using the factored forms:
\[
\frac{(x + 3)(x - 1)}{(x + 4)(x - 1)} \cdot \frac{(x + 2)(x + 4)}{(x + 3)(x - 3)}.
\]
**Step 3: Simplify the expression.**
Now, let's multiply the fractions:
\[
\frac{(x + 3)(x - 1)(x + 2)(x + 4)}{(x + 4)(x - 1)(x + 3)(x - 3)}.
\]
Now we can cancel common factors in the numerator and denominator:
- \(x + 3\) cancels with \(x + 3\).
- \(x - 1\) cancels with \(x - 1\).
- \(x + 4\) cancels with \(x + 4\).
This leaves us with:
\[
\frac{(x + 2)}{(x - 3)}.
\]
Thus, the expression simplifies to:
\[
\frac{x + 2}{x - 3}.
\]
**Final Result:**
\(\frac{x + 2}{x - 3}\) is the simplified form of the original product.
The correct response is:
\[
\frac{x + 2}{x - 3}
\]
\[
\frac{x^2 + 2x - 3}{x^2 + 3x - 4} \cdot \frac{x^2 + 6x + 8}{x^2 - 9},
\]
we will simplify each rational expression before multiplying.
**Step 1: Factor each polynomial.**
1. For \(x^2 + 2x - 3\):
\[
x^2 + 2x - 3 = (x + 3)(x - 1)
\]
2. For \(x^2 + 3x - 4\):
\[
x^2 + 3x - 4 = (x + 4)(x - 1)
\]
3. For \(x^2 + 6x + 8\):
\[
x^2 + 6x + 8 = (x + 2)(x + 4)
\]
4. For \(x^2 - 9\) (difference of squares):
\[
x^2 - 9 = (x + 3)(x - 3)
\]
**Step 2: Substitute the factors back into the expression.**
Now we rewrite the expression using the factored forms:
\[
\frac{(x + 3)(x - 1)}{(x + 4)(x - 1)} \cdot \frac{(x + 2)(x + 4)}{(x + 3)(x - 3)}.
\]
**Step 3: Simplify the expression.**
Now, let's multiply the fractions:
\[
\frac{(x + 3)(x - 1)(x + 2)(x + 4)}{(x + 4)(x - 1)(x + 3)(x - 3)}.
\]
Now we can cancel common factors in the numerator and denominator:
- \(x + 3\) cancels with \(x + 3\).
- \(x - 1\) cancels with \(x - 1\).
- \(x + 4\) cancels with \(x + 4\).
This leaves us with:
\[
\frac{(x + 2)}{(x - 3)}.
\]
Thus, the expression simplifies to:
\[
\frac{x + 2}{x - 3}.
\]
**Final Result:**
\(\frac{x + 2}{x - 3}\) is the simplified form of the original product.
The correct response is:
\[
\frac{x + 2}{x - 3}
\]