Question

After fully simplifying the expression

[ (2x^2 + 3x + 1) / (x^2 + 1x - 6) ] / [ (x^2 + 2x +1) / 2x^2 - 5x + 2) ]

Answer 1

Factor those polynomials, then we'll talk ;p

Answer 2

LOL, okay. good thing I already have it on paper.

Answer 3

[ (2x+1)(x+1) / (x+6)(x-1) ] / [ (x+1)(x+1) / (2x^2-2)(x-2) ]

Answer 4

That's what I have for factoring, i didn't flip anything yet..

Answer 5

Very close.. I disagree with the (x+6)(x-1) denominator you have..

\[\huge\qquad\frac{\frac{(2x+1)(x+1)}{(x+3)(x-2)}}{\frac{(x+1)^2}{(2x-1)(x-2)}}\]

Answer 6

Good so far?

Answer 7

OH! okay, I see now

Answer 8

Let me try to figure it out and you can check after me, please! Thanks!

Answer 9

Certainly.

Answer 10

Okay, so I got..

(2x+1)(2x-1)
------------
(x+3)(x+1)

Answer 11

Yep, I got that too ..

\[\large\qquad \qquad \frac{\frac{(2x+1)(x+1)}{(x+3)(x-2)}}{\frac{(x+1)^2}{(2x-1)(x-2)}}\]\[\large= \qquad \frac{\frac{(2x+1)(x+1)}{\cancel{(x+3)(x-2)}}}{\frac{(x+1)^2}{\cancel{(2x-1)(x-2)}}} \cdot \frac{\cancel{(x-2)(x+3)}(2x-1)}{\cancel{(x-2)}(x+3)\cancel{(2x-1)}}\]\[\large= \qquad \frac{(2x+1)\cancel{(x+1)}(2x-1)}{(x+1)^\cancel{2}(x+3)}\]\[\large= \qquad \frac{(2x+1)(2x-1)}{(x+1)(x+3)}\]\[\large= \qquad \frac{4x^2 - 1}{x^2 + 4x + 3}\]

Answer 12

You are miracles. Thank you so much.

Answer 13

Nice work!