Factor

[hopelynnwelch]

I am taking Math 262 after being out of math for a really long time. I might be hanging myself by doing this but I tested into the course just barely. For these reasons I plan to work extra hard this semester. I used this forum a long time ago and it was extremely useful for checking myself and so I know it will benefit me greatly this semester. Thanks in advance for any help!

Here is the problem:

Factor [ 3(x+2)^2 - 6(x+2)^3 ] / [ 2(x+2)^4 ]

Here is what I got:

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

I am wondering if I should cancel or multiply out as part of "factoring" since all terms are separated by a multiply? Or leave my answer as it is above?

 

[Ishuda]

I am taking Math 262 after being out of math for a really long time. I might be hanging myself by doing this but I tested into the course just barely. For these reasons I plan to work extra hard this semester. I used this forum a long time ago and it was extremely useful for checking myself and so I know it will benefit me greatly this semester. Thanks in advance for any help!

Here is the problem:

Factor [ 3(x+2)^2 - 6(x+2)^3 ] / [ 2(x+2)^4 ]

Here is what I got:

[ 3(x+2)^2 ( 1 + (-2x-4) ) ] / [ 2(x+2)^4 ]

I am wondering if I should cancel or multiply out as part of "factoring" since all terms are separated by a multiply? Or leave my answer as it is above?

Starting with the (corrected) answer above we have
[ 3(x+2)² ( 1 + (-2x - 4) ] / [ 2(x+2)⁴ ] = [ - 3(x+2)² (2x + 3) ] / [ 2(x+2)⁴ ]
You can now cancel the (x+2)² in both numerator and denominator (as part of the (x+2)⁴) to get
[ -3 (2x + 3) ] / [ 2(x+2)² ]
or, if you like,
\(\displaystyle -\frac{3}{2} \frac{2x + 3}{(x+2)^2}\)

 

[Steven G]

I am taking Math 262 after being out of math for a really long time. I might be hanging myself by doing this but I tested into the course just barely. For these reasons I plan to work extra hard this semester. I used this forum a long time ago and it was extremely useful for checking myself and so I know it will benefit me greatly this semester. Thanks in advance for any help!

Here is the problem:

Factor [ 3(x+2)^2 - 6(x+2)^3 ] / [ 2(x+2)^4 ]

Here is what I got:

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

I am wondering if I should cancel or multiply out as part of "factoring" since all terms are separated by a multiply? Or leave my answer as it is above?

maybe letting u=(x+2) will make things easier (or if you feel comfortable, let u=(x+2)^2). So your problem becomes [3u^2 - 6u^3]/2u^4. The two terms in the numerator have factors of 3 AND u^2 in both, so we factor them out.

We get 3u^2(1-2u)/2u^4. The numerator is 3*u*u(3-2u) while the denominator is 2*u*u*u*u. Clearly we can cancel out two u's in the top with two u's in the denominator giving us

3(1-2u)/(2u^2). Now put (x+2) in place of u and clean up your answer.

Then show us your work.

 

[hopelynnwelch]

Starting with the (corrected) answer above we have
[ 3(x+2)² ( 1 + (-2x - 4) ] / [ 2(x+2)⁴ ] = [ - 3(x+2)² (2x + 3) ] / [ 2(x+2)⁴ ]
You can now cancel the (x+2)² in both numerator and denominator (as part of the (x+2)⁴) to get
[ -3 (2x + 3) ] / [ 2(x+2)² ]
or, if you like,
\(\displaystyle -\frac{3}{2} \frac{2x + 3}{(x+2)^2}\)

Ohh I dropped the 1... Yikes I am rusty. Hopefully this all comes back to me soon. Thank you!

 

[hopelynnwelch]

maybe letting u=(x+2) (or if you feel comfortable, let u=(x+2)^2). So your problem becomes [3u^2 - 6u^3]/2u^4. The two terms in the numerator have a factor of u^2 in both, so we factor it out.

We get u^2(3-2u)/2u^4. The numerator is u*u(3-2u) while the denominator is 2*u*u*u*u. Clearly we can cancel out two u's in the top with two u's in the denominator giving us

(3-2u)/u^2. Now put (x+2) in place of u and clean up your answer.

Then show us your work.

And substituting u is great! I forgot I could do that! Thank you!