To solve the given problem, let's break it down step-by-step.
The original expression is:
[tex]\[
\frac{2x^2 - 5x - 3}{4x^2 + 12x + 5} \div \frac{3x^2 - 11x + 6}{6x^2 + 11x - 10}
\][/tex]
To simplify this, we multiply the first fraction by the reciprocal of the second fraction:
[tex]\[
\frac{2x^2 - 5x - 3}{4x^2 + 12x + 5} \cdot \frac{6x^2 + 11x - 10}{3x^2 - 11x + 6}
\][/tex]
Next, we factor all the polynomials in the numerators and denominators:
1. [tex]\(2x^2 - 5x - 3\)[/tex] factors to [tex]\((x - 3)(2x + 1)\)[/tex].
2. [tex]\(4x^2 + 12x + 5\)[/tex] factors to [tex]\((2x + 5)(2x + 1)\)[/tex].
3. [tex]\(3x^2 - 11x + 6\)[/tex] factors to [tex]\((x - 3)(3x - 2)\)[/tex].
4. [tex]\(6x^2 + 11x - 10\)[/tex] factors to [tex]\((2x + 5)(3x - 2)\)[/tex].
Now substitute these factored forms back into the expression:
[tex]\[
\frac{(x - 3)(2x + 1)}{(2x + 5)(2x + 1)} \cdot \frac{(2x + 5)(3x - 2)}{(x - 3)(3x - 2)}
\][/tex]
We can see which factors are common and can be canceled out:
- [tex]\( (x - 3) \)[/tex] in the numerator of the first fraction and the denominator of the second fraction.
- [tex]\( (2x + 1) \)[/tex] in the numerator and the denominator of the first fraction.
- [tex]\( (2x + 5) \)[/tex] in the numerator of the second fraction and the denominator of the first fraction.
- [tex]\( (3x - 2) \)[/tex] in the numerator and the denominator of the second fraction.
By canceling out these common factors, we are left with:
[tex]\[
1 \cdot 1 = 1
\][/tex]
Hence, the correct multiplication expression that matches this simplification is:
[tex]\[
\frac{(x - 3)(2x + 1)}{(2x + 1)(2 x + 5)} \cdot \frac{(2 x+ 5)(3 x - 2)}{(x - 3)(3 x - 2)}
\][/tex]
The equivalence confirms that the correct multiplication expression is:
[tex]\[
\frac{(x-3)(2 x+1)}{(2 x+1)(2 x+5)} \cdot \frac{(2 x+5)(3 x-2)}{(x-3)(3 x-2)}
\][/tex]
Therefore, the simplified result after canceling common factors indicates the correct answer is:[tex]\(3\)[/tex]:
[tex]\[
3
\][/tex]
