How do I factor out negative signs in equations?

[sobergeek23]

i have no idea how to use the punctuation marks for writing equations..ill do my best..

(3a+4b)(a-3b) over b(3a+4b) multiplied by 1 over b^2(3b-a)(3b+a)

during the cross cancelling the 3b-a would be multiplied by a negative one to switch the problem over right? (-1)(3b-a) in order to cross cancel with the (a-3b) in the top left..
the answer was 1 over b^3(a-3b) but i don't understand the whole negative sign thing..isnt (3b-a) the same as (-a+3b)? so factoring out the negative would make it (a-3b) and it cancels with the top left fraction but then i also have the (b) (b^2)(3b+a) on the bottom so how do they get b^3(a-3b)? i get the b^3 part ..i thought maybe the negative one i factored out in the (3b-a) would have also applied to the (3b+a) but then wouldn't it be (-3b-a) ..im so confussed on factoring out negatives..

 

[fresh_42]

i have no idea how to use the punctuation marks for writing equations..ill do my best..

(3a+4b)(a-3b) over b(3a+4b) multiplied by 1 over b^2(3b-a)(3b+a)

during the cross cancelling the 3b-a would be multiplied by a negative one to switch the problem over right? (-1)(3b-a) in order to cross cancel with the (a-3b) in the top left..
the answer was 1 over b^3(a-3b)

Sure there isn't a typo? What happened to $(3b+a)$?

but i don't understand the whole negative sign thing..isnt (3b-a) the same as (-a+3b)? so factoring out the negative would make it (a-3b) and it cancels with the top left fraction but then i also have the (b) (b^2)(3b+a) on the bottom so how do they get b^3(a-3b)?

see above

i get the b^3 part ..i thought maybe the negative one i factored out in the (3b-a) would have also applied to the (3b+a) but then wouldn't it be (-3b-a) ..im so confussed on factoring out negatives..

I don't really see where the $-1$ has gone to, neither in the quoted answer nor in the confusion of your reasoning.

To write formulas, you should read
https://www.physicsforums.com/help/latexhelp/

It's not that difficult but makes reading a lot easier. I think you have all you need to simplify the quotient. Just keep track of your parts. A simple trick to see, whether a step is write or wrong, is to plug in some small numbers for $a$ and $b$, like $\pm 1$ or $\pm 2$.

 

[ehild]

i have no idea how to use the punctuation marks for writing equations..ill do my best..

(3a+4b)(a-3b) over b(3a+4b) multiplied by 1 over b^2(3b-a)(3b+a)

during the cross cancelling the 3b-a would be multiplied by a negative one to switch the problem over right? (-1)(3b-a) in order to cross cancel with the (a-3b) in the top left..
the answer was 1 over b^3(a-3b) but i don't understand the whole negative sign thing..isnt (3b-a) the same as (-a+3b)? so factoring out the negative would make it (a-3b) and it cancels with the top left fraction but then i also have the (b) (b^2)(3b+a) on the bottom so how do they get b^3(a-3b)? i get the b^3 part ..i thought maybe the negative one i factored out in the (3b-a) would have also applied to the (3b+a) but then wouldn't it be (-3b-a) ..im so confussed on factoring out negatives..

Was it the problem?
$$\frac{(3a+4b)(a-3b)}{b(3a+4b)}\frac{1}{b^2(3b-a)(3b+a)}$$
$$=\frac{(3a+4b)(a-3b)}{b(3a+4b)}\frac{1}{-b^2(a-3b)(3b+a)}$$

You canceled out a-3b, and you got a (-1) factor in the denominator. Dividing by -1 makes the sign of the whole fraction negative, so it is
$$-\left(\frac{(3a+4b)}{b(3a+4b)}\frac{1}{b^2(3b+a)}\right)$$after further simplification, you get the desired formula.

The minus sign is the same as a factor of (-1).
Remember the rules when calculating with factors. $$\frac{a}{bc}=\frac{1}{b}\frac{a}{c}$$
If b = -1,
$$\frac{a}{-c}=\frac{1}{-1}\frac{a}{c}$$
and you know that $$\frac{1}{-1}= -1$$
so $$\frac{a}{-c}=-\frac{a}{c}$$

 

[sobergeek23]

#3 ..writing it out not that hard? looking at that page just made my eyes cross and my brain short circuit..ugh forget it..

 

[Ray Vickson]

#3 ..writing it out not that hard? looking at that page just made my eyes cross and my brain short circuit..ugh forget it..

Can you write (3a+4b)(a-3b)/( b(3a+4b))? Is that easier than writing \frac{(3a+4b)(s-3b)}{b(3a+4)}? Basically, that's all there is to it!

When you write x = \frac{a}{b} in LaTeX, you get $x = \frac{a}{b}$ after you enclose everything between two # symbols, like this: # # your stuff # # (remove spaces between the two # signs at the start and the end). Using # delimiters produces an "in-line" formula. If you want a "displayed" formula/equation, like this
$$ x =\frac{a}{b},$$
you should replace the # symbols by $ signs, so write $ $ your stuff $ $ (with no spaces between the two $ signs at the start and the end).

 

[sobergeek23]

no idea what that means...the first thing u wrote made sense, the rest way over my head..this site won't let me post pictures either otherwise id just upload a pic of the problem