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The quotient of

[tex]\[
\frac{5n^2(m^2 + 2m - 3)}{10n(m^2 - 3m + 2)} \div \frac{n^2(m^2 - 3m - 18)}{4n^2(m^2 - 8m + 12)}
\][/tex]

is

[tex]\(\boxed{\phantom{answer}}\)[/tex]



Answer :

To find the quotient of the given expression, we can break the problem down into several steps.

We need to divide the first fraction by the second fraction:

[tex]\[ \frac{\frac{5 n^2 (m^2 + 2m - 3)}{10 n (m^2 - 3m + 2)}}{\frac{n^2 (m^2 - 3m - 18)}{4 n^2 (m^2 - 8m + 12)}} \][/tex]

### Step 1: Rewrite Division as Multiplication by the Reciprocal
First, let's rewrite this division problem as a multiplication problem by taking the reciprocal of the second fraction:

[tex]\[ \frac{5 n^2 (m^2 + 2m - 3)}{10 n (m^2 - 3m + 2)} \times \frac{4 n^2 (m^2 - 8m + 12)}{n^2 (m^2 - 3m - 18)} \][/tex]

### Step 2: Factor All Quadratics
Now let's factor all quadratic expressions.

- [tex]\( m^2 + 2m - 3 \)[/tex] factors to [tex]\((m + 3)(m - 1)\)[/tex]
- [tex]\( m^2 - 3m + 2 \)[/tex] factors to [tex]\((m - 1)(m - 2)\)[/tex]
- [tex]\( m^2 - 3m - 18 \)[/tex] factors to [tex]\((m + 3)(m - 6)\)[/tex]
- [tex]\( m^2 - 8m + 12 \)[/tex] factors to [tex]\((m - 6)(m - 2)\)[/tex]

Substitute these factorizations back in:

[tex]\[ \frac{5 n^2 (m+3)(m-1)}{10 n (m-1)(m-2)} \times \frac{4 n^2 (m-6)(m-2)}{n^2 (m+3)(m-6)} \][/tex]

### Step 3: Simplify the Expression
Now we will cancel out common factors in the numerators and denominators:

[tex]\[ \frac{5 n^2 (m+3)(m-1)}{10 n (m-1)(m-2)} \times \frac{4 n^2 (m-6)(m-2)}{n^2 (m+3)(m-6)} \][/tex]

Here, [tex]\((m+3)\)[/tex], [tex]\((m-1)\)[/tex], [tex]\((m-2)\)[/tex], and [tex]\((m-6)\)[/tex] all appear in both the numerator and the denominator, so they cancel each other out. Thus, we're left with:

[tex]\[ \frac{5 n^2}{10 n} \times \frac{4 n^2}{n^2} \][/tex]

[tex]\[ = \frac{20 n^4}{10 n^3} \][/tex]

Simplify this:

[tex]\[ = 2 n^3 \][/tex]

Therefore, the quotient is:

[tex]\[ \boxed{2 n^3} \][/tex]