Question

This is algebra 2. Would like an answer ASAP. Thanks! Between the two chunks of numbers, thats a division sign, not an addition sign.

Answer 1

Answer:

Simplified form of $\frac{x+1}{x^2+x-6}\div \frac{x^2+5x+4}{x-2}$ is $\mathbf{\frac{1}{(x+3)(x+4)} }$

Option A is correct answer.

Step-by-step explanation:

We need to find simplified form of $\frac{x+1}{x^2+x-6}\div \frac{x^2+5x+4}{x-2}$

Solving the given expression:

$\frac{x+1}{x^2+x-6}\div \frac{x^2+5x+4}{x-2}$

First we find factors of $x^2+x-6$

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

So, factors of $x^2+x-6$ are $(x-2)(x+3)$

Now, fining the factors of $x^2+5x+4$

$x^2+5x+4\\=x^2+4x+x+4\\=x(x+4)+1(x+4)\\=(x+1)(x+4)$

So, factors of $x^2+5x+4$ are $(x+1)(x+4)$

Now the given expression will become:

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

Now, converting division sign into multiplication sign:

$=\frac{x+1}{(x-2)(x+3)}\times \frac{x-2} {(x+1)(x+4)}\\Cancelling, \:common\:terms\\=\frac{1}{(x+3)(x+4)}$

So, simplified form of $\frac{x+1}{x^2+x-6}\div \frac{x^2+5x+4}{x-2}$ is $\mathbf{\frac{1}{(x+3)(x+4)} }$

Option A is correct answer.