Question

Please help!
Simplify: (4x^2-14x-30)/(x^2-1) ÷ (2x^2+5x+3)/(x^2-6x+5)

Answer 1

Please use parentheses when posting the question to separate the numerator from the denominator.

Answer 2

Sorry.

Answer 3

That's a lot better... if you don't do that, then it leaves your expressions open to unnecessary interpretation.

Answer 4

Had to fix 4x^2-14x-30)/(x^2-1)

Answer 5

Sorry.

Answer 6

So, now we have

\[\frac{4x^2 - 14x - 30}{x^2} \div \frac{2x^2 + 5x + 3}{x^2 - 6x + 5}\]
Which can be re-written as

\[\frac{4x^2 - 14x - 30}{x^2} \times \frac{x^2 - 6x + 5 }{2x^2 + 5x + 3}\]

The next thing to do is factor, which afterwards you get:
\[\frac{2(2x + 3)(x -5)}{x^2} \times \frac{(x - 1)(x- 5) }{(x + 1)(2x + 3)}\]

Answer 7

From here, you cancel factors of one

Answer 8

How did you figure out the 2(2x +3) (x-5)?

Answer 9

4x^2 - 14x - 30

Factor out 2 which is common to all 3 terms:
2(2x^2 - 7x - 15)

Next factor x^2 - 7x - 15:
2(2x^2 - 10x + 3x - 15)
2(2x(x - 5) + 3(x - 5))
2((x - 5)(2x + 3))
2(x - 5)(2x + 3)

Answer 10

Thanks.

Answer 11

Are you sure you understood it?

Answer 12

That part. I am still working on the problem.

Answer 13

Suppose you were given another quadratic to factor. Do you think you could factor it in the same manner?

Answer 14

Maybe.

Answer 15

Why is it (x+1) (2x+3) and not (x+3) (x+2)?

Answer 16

Multiply (x + 3)(x + 2) to find out

Answer 17

Use the distributive rule when multiplying a(b + c) = ab + ac

Similarly:
(x + 3)(x + 2) = x(x + 3) + 2(x + 3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6

Answer 18

\[2x^2 + 5x + 3 \ne x^2 + 5x + 6\]