Multiplying Rational expressions

[silverdragon316]

I would like to know if I am doing this correctly.
The problem is: I have to multiply

(2x^2-7x-15) (x-4)
(x^2-6x+8) (2x+3)

I then factored it and got:

2(x-5)(x+3)(x-4)
2(x-2)(x-4)(x-3)

then I cancelled out and got:

x-5
x-2 for the answer.[/u]

 

[jonboy]

Hrm I'm kinda confused on your notation.

Do you mean this: \(\displaystyle \L \;\frac{2x^2\,-\,7x\,-\,15\,*\,x\,-\,4 }{x^2\,-\,6x\,+\,8 2x\,+\,3}\)

Or perhaps: \(\displaystyle \L \;2x^2\,-\,7x\,-\,15\,*\,x\,-\,4\,\cdot\,x^2\,-\,6x\,+\,8 2x\,+\,3\)

 

[morson]

And that's a correct answer so long as x isn't one of 4, 2, or -1.5; otherwise this would be meaningless.

 

[Denis]

2x^2-7x-15 * x-4
x^2-6x+8 2x+3

Is that a division? If so:
(2x^2 - 7x - 15)(x - 4)
===============
(x^2 - 6x + 8)(2x + 3)


I then factored it and got:
2(x-5)(x+3)(x-4)
2(x-2)(x-4)(x-3)

No. 2x^2 - 7x - 15 = (x - 5)(2x + 3); so numerator = (x - 5)(2x + 3)(x - 4)
Denominator also incorrect: I'll let YOU fix it :shock:

 

[morson]

2x^2-7x-15 * x-4
x^2-6x+8 2x+3

Is that a division? If so:
(2x^2 - 7x - 15)(x - 4)
===============
(x^2 - 6x + 8)(2x + 3)


I then factored it and got:
2(x-5)(x+3)(x-4)
2(x-2)(x-4)(x-3)

No. 2x^2 - 7x - 15 = (x - 5)(2x + 3); so numerator = (x - 5)(2x + 3)(x - 4)
Denominator also incorrect: I'll let YOU fix it :shock:

Coincidentally, they got the right answer, despite the method being completely incorrect. I guess in this case two wrongs make a right.

 

[stapel]

...two wrongs make a right.

Two wrongs do not make a right. But two Wrights make an airplane. :wink:

Eliz.