Question

How do you multiply `(6x^2 + x - 12 )/(2x^2 - 5x - 12 )*(3x^2 - 14x + 8)/( 9x^2 - 18x + 8)`?

Answer 1

You can factorize all these quadratic equations first.

In order to factorize them, all tou need to do is find its roots and then turn each root into a factor by equaling it to zero.

For example, let's say you've found `x=1/2`. Then, your factor will be `(2x-1)=0`.

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`((3x-4)(2x+3))/((2x+3)(x-4))*((3x-2)(x-4))/((3x-2)(3x-4))`

Now, we can proceed to cancel some binomials that are both dividing and multiplying this equation.

`(cancel(3x-4)(2x+3))/((2x+3)(x-4))*((3x-2)(x-4))/((3x-2)cancel(3x-4))`

`(cancel(3x-4)cancel(2x+3))/(cancel(2x+3)(x-4))*((3x-2)(x-4))/((3x-2)cancel(3x-4))`

`(cancel(3x-4)cancel(2x+3))/(cancel(2x+3)cancel(x-4))*((3x-2)cancel(x-4))/((3x-2)cancel(3x-4))`

`(cancel(3x-4)cancel(2x+3))/(cancel(2x+3)cancel(x-4))*(cancel(3x-2)cancel(x-4))/(cancel(3x-2)cancel(3x-4))`

Thus, your result is `1`. :)

Answer 2

It may help to factor each of the quadratics into linear factors first - There may be some terms we can eliminate :-)

`6x^2+x-12 = (3x-4)(2x+3)`
`3x^2-14x+8 = (3x-2)(x-4)`
`2x^2-5x-12 = (2x+3)(x-4)`
`9x^2-18x+8 = (3x-4)(3x-2)`

So

`(6x^2+x-12)/(2x^2-5x-12)*(3x^2-14x+8)/(9x^2-18x+8)`

`=((3x-4)(2x+3))/((2x+3)(x-4))*((3x-2)(x-4))/((3x-4)(3x-2))`

`=(3x-4)/(x-4)*(x-4)/(3x-4)`

`=1`

Note that this equation is only valid when the denominator is non-zero, which will be when `x` is not one of the values:
`-3/2`, `4`, `4/3` or `2/3`.