Math geniuses I really can use your help please I have to reduce the following?

Question

I have a final and I am scared I will not get my credit if I fail:( Can you please explain each step as you go along?

6x^2+5x-6 * 12x^2-x-6 / 12x^2-17x+6
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12x-3x^2 6x^2+13x+6 2x^2-8X

PLEASE HELP

Answer 1

It's hard to see exactly what the problem is.

The general principal is to:
1. Multiply out the top and bottom and group together like terms.
2. Try to factor the resulting equations on both top and bottom.
3. Cancel out any common factors.

Hmm no I think you want to try to factor the top and bottom equations separately first.

For example, 6x^2 + 5x - 6
I worked out that this is (2x + 3)(3x -2) by doing the following:

(2x + a)(3x + b) => 6x^2 + (3b + 2a)x + ab

3a + 2b = 5
ab = -6
a = 3, b = -2

Similarly, 12x^2 - x - 6 is (3x + 2)(4x - 3)
I guessed it was (3x + a)(4x + b) and worked from there. If that didn't work out, I was going to try 2,6 and 1,12

And similarly, 12x^2 - 17x + 6 is (3x -2)(4x -3)

On the bottom,
12x - 3x^2 => -3x(x - 4)

2x^2 - 8x => 2x(x - 4)

6x^2 + 13x + 6 => (3x + 2)(2x + 3)

So we have:
[ (2x + 3)(3x -2) (3x + 2)(4x - 3) / (3x -2)(4x -3) ] /
[ -6x^2(x - 4)(x - 4)(3x + 2)(2x +3) ]

=> (2x + 3)(3x + 2) / [ -6x^2 (x-4)^2 (3x + 2)(2x + 3) ]

=> -1 / 6x^2 ( x - 4)^2

Answer 2

I assume this is meant to be
[(6x^2 + 5x - 6) / (12x - 3x^2)] * [(12x^2 - x - 6) / (6x^2 + 13x + 6)] / [(12x^2 - 17x + 6) / (2x^2 - 8x)]
Factorising and inverting the final fraction to change the division into a multiplication gives
= [(3x-2)(2x+3) / (3x(4x-1))] * [(4x-3)(3x+2) / ((2x+3)(3x+2))] * [2x(x-4) / ((4x-3)(3x-2))]
Cancelling common factors within and across terms gives us
= [1 / (3(4x-1))] * 1 * [2(x-4) / 1]
= 2(x-4) / 3(4x-1)

Answer 3

group you terms using parenthesis.. +6 2x^2-8x