factoring/simplification question

markraz

markraz

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Regarding the simplification of this fraction .

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


The answer to this is 3/(2x+1)^2


why is it that I can't first cancel the (4x+2) in the numerator with one of the (4x+2)^2 in denominator?
how do I know that I'm actually suppose to factor the denominator first?? [2(2x+1)]^2
and then reduce the 6 in the numerator with the factored 2^2 in the denominator?

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

is there some order of specific operations I need to follow when simplifying?
What is the standard procedure? How do you know what to do first?
do you always factor first before canceling/reducing? is this part of that pemdas business?

thanks
 
Last edited:
markraz said:
Regarding the simplification of this fraction .

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


The answer to this is 3/(2x+1)^2


why is it that I can't first cancel the (4x+2) in the numerator with one of the (4x+2)^2 in denominator?

The whole numerator would have to be one product, (4x + 2) being one of the factors.

how do I know that I'm actually suppose to factor the denominator first?? [2(2x+1)]^2
and then reduce the 6 in the numerator with the factored 2^2 in the denominator?

-> 6/4(2x+1)^2 \(\displaystyle \ \ \ \ \)You must have grouping symbols around the denominator: 6/[4(2x+1)^2]
Click to expand...
.
 
markraz said:
[(4x+2)(3)-(3x)(4)] / [(4x+2)^2]

The answer to this is 3/(2x+1)^2

why is it that I can't first cancel the (4x+2) in the numerator with one of the (4x+2)^2 in denominator?
Click to expand...
Because you can only cancel factors, not terms. You can't cancel the 4x + 2 out of PART of the numerator for the same reason that you can't cancel the 2s in 12/23 to get 1/3: 12 = 10 + 2 and 23 = 2*10 + 3, but there is no common factor to be cancelled. ;)
 
thanks, what specific arithmetic or algebra rule is this dictated by?
 
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