Question

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Answer 1

Option 3

The solution for given expression is $\frac{(x - 4)(x - 4)}{(x + 3)(x + 1)}$

Solution:

Given that we have to divide,

$\frac{x^2 -16}{x^2 + 5x + 6} \div \frac{x^2 + 5x + 4}{x^2 -2x - 8}$   ---- (A)

Let us first factorize each term and then solve the sum

Using $a^2 - b^2 = (a + b)(a - b)$

$x^2 -16 = x^2 - 4^2 = (x + 4)(x -4)$  ----- (1)

$x^2 + 5x + 6 = (x + 2)(x + 3)$   ----- (2)

$x^2 + 5x + 4 = (x+1)(x + 4)$   ---- (3)

$x^2 -2x - 8 = (x-4)(x + 2)$   ---- (4)

Now substituting (1), (2), (3), (4) in (A) we get,

$\frac{(x + 4)(x -4)}{(x +2)(x +3)} \div \frac{(x+1)(x+4)}{(x-4)(x +2)}$

To do division with fractions, we turn the second fraction upside down and change the division symbol to a multiplication symbol at the same time. Then we treat this as a multiplication problem, by multiplying the numerators and the denominators separately.

$\frac{(x + 4)(x -4)}{(x +2)(x +3)} \times \frac{(x - 4)(x + 2)}{(x + 1)(x + 4)}$

On cancelling terms we get,

$= \frac{(x -4)(x-4)}{(x + 3)(x + 1)}$

Thus option 3 is correct