15x^2 -25x -10. Answer A:(5x-10)(3x+1).Answer B:5(3x+1)(x-2).

Is one more right than the other? Or is A wrong simply due to method?

Thanks.

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- Jul 16th 2010, 12:48 PMIngersollAre both answers right?
**15x^2 -25x -10**. Answer A:*(5x-10)(3x+1).*Answer B:*5(3x+1)(x-2).*

Is one more right than the other? Or is A wrong simply due to method?

Thanks. - Jul 16th 2010, 12:55 PMeumyang
Depends on what you mean by "right." If the question is to factor

**completely**, then Answer B would be better, because in the first factor of Answer A (5x - 10), you can factor out the greatest common monomial factor, which is 5. So

(5x - 10)(3x + 1) =

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

... which is Answer B with the binomials swapped. - Jul 16th 2010, 01:44 PMIngersoll
Gotcha. Thank you very much. That clears things up for me. :)

- Jul 16th 2010, 02:17 PMArchie Meade
They are not really "answers".

You have three alternative ways to write the same expression.

$\displaystyle 15x^2-25x-10$

factor the 15 and the -10 to get

$\displaystyle (5x-10)(3x+1)$

factor the 5 and -10 in the left factor to get

$\displaystyle 5(x-2)(3x+1)$

$\displaystyle 15x^2-25x-10=(5x-10)(3x+1)=5(x-2)(3x+1)$

If you choose any value for x, you will get the same result if you place it in any of the 3 equivalent versions of the expression.

If however, you began from

$\displaystyle 15x^2=25x+10$

and you want to "solve" for x, then both sides are equal, so subtract them and the answer is zero

$\displaystyle 15x^2-25x-10=0$

It's not obvious what x is yet...

$\displaystyle (5x-10)(3x+1)=0$

Now it's much clearer what x is

$\displaystyle 5(x-2)(3x+1)=0$

Clearly x=2 is a solution since if x=2, x-2=0 and 0(anything)=0.

One final step...

$\displaystyle 5(x-2)3\left(x+\frac{1}{3}\right)=0$

Now we can see that $\displaystyle x=-\frac{1}{3}$ is also a solution.

That is the advantage to factoring.