unimatix
  • unimatix
Simplify Algebra (in comments)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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unimatix
  • unimatix
\[4(-8)(2-8x)^3(x^2-9)^3+3(2x)(2-8x)^4(x^2-9)^2\]
unimatix
  • unimatix
The next step shown looks like \[4(2-8x)^3(x^2-9)^2(20x^2-3x-72)\]
unimatix
  • unimatix
I'm having trouble understand how that was arrived at.

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anonymous
  • anonymous
hold on solving them
unimatix
  • unimatix
Okay, thanks!
anonymous
  • anonymous
alright here it is, use this website to do the rest http://prntscr.com/8nfl0b
anonymous
  • anonymous
(www.Mathway.com)
unimatix
  • unimatix
wut?
Nnesha
  • Nnesha
:=)
Nnesha
  • Nnesha
heya
Nnesha
  • Nnesha
take out the common factor
unimatix
  • unimatix
I just need it simplified to the second step.
unimatix
  • unimatix
because then I am solving equal to 0.
unimatix
  • unimatix
that's where my confusion is
Nnesha
  • Nnesha
\[4(-8)\color{Red}{(2-8x)^3}\color{blue}{(x^2-9)^3}+3(2x)\color{Red}{(2-8x)^4}\color{blue}{(x^2-9)^2}\] see there is GCF (greatest common factor
Nnesha
  • Nnesha
|dw:1443912816495:dw| now what is common in both terms sorry about the writing ....:/\:
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @unimatix The next step shown looks like \[4(2-8x)^3(x^2-9)^2(20x^2-3x-72)\] \(\color{blue}{\text{End of Quote}}\) i think you forgot the negative sign it's `-4`
Nnesha
  • Nnesha
\[4(-8)\color{Red}{(2-8x)^3}\color{blue}{(x^2-9)^3}+3(2x)\color{Red}{(2-8x)^4}\color{blue}{(x^2-9)^2}\] (x^2-9)^3 can be wriitten as (x^2-9)^2(x^2-9) \[4(-8)\color{Red}{(2-8x)^3}\color{blue}{(x^2-9)^2(x^2-9)}+3(2x)\color{Red}{(2-8x)^4}\color{blue}{(x^2-9)^2}\] so as you can (x^2-9)^2 is common
Nnesha
  • Nnesha
ohhh well you're not online can't work without u
mathstudent55
  • mathstudent55
\(4(-8)(2-8x)^3(x^2-9)^3+3(2x)(2-8x)^4(x^2-9)^2\) \(=-32(2-8x)^3(x^2-9)^3+3(2x)(2-8x)^4(x^2-9)^2\) \(=(2 - 8x)^3(x^2 - 9)^2[-32(x^2-9)+3(2x)(2-8x)]\) \(=(2 - 8x)^3(x^2 - 9)^2[-32x^2+288+12x - 48x^2]\) \(=(2 - 8x)^3(x^2 - 9)^2(-80x^2+12x + 288)\)
unimatix
  • unimatix
3rd line down I'm confused
unimatix
  • unimatix
How are you moving the -32?
mathstudent55
  • mathstudent55
\(=(2 - 8x)^3(x^2 - 9)^2(-80x^2+12x + 288)\) \(= -4 (2 - 8x)^3(x^2 - 9)^2(20x^2 -3 x - 72)\)
mathstudent55
  • mathstudent55
Here is the second line. You're ok with it? \(=-32(2-8x)^3(x^2-9)^3+3(2x)(2-8x)^4(x^2-9)^2\)
unimatix
  • unimatix
Yeah!
mathstudent55
  • mathstudent55
From the 2nd line to the third line, we factor out two factors: \(=-32\color{red}{(2-8x)^3}\color{purple}{(x^2-9)^3}+3(2x)\color{red}{(2-8x)^4}\color{purple}{(x^2-9)^2}\)
mathstudent55
  • mathstudent55
We have (in red) the factor 2 - 8x. It is to the 3 power in the first part, and to the 4th power in the second part. We can factor out (8 - 2x)^3. Then (in purple) we have the factor x^2 - 9 to the 3rd power and to the 2th power. We can factor out (x^2 - 9)^2. Ok so far?
unimatix
  • unimatix
OKay I understand that step
unimatix
  • unimatix
good on the 4th line too
unimatix
  • unimatix
Okay I got it.
mathstudent55
  • mathstudent55
Here is the 3rd line after factoring out those two factors: \(=\color{red}{(2 - 8x)^3}\color{purple}{(x^2 - 9)^2}[-32\color{purple}{(x^2-9)}+3(2x)\color{red}{(2-8x)}]\)
mathstudent55
  • mathstudent55
Ok.
unimatix
  • unimatix
I'm not really used to factoring out those factors I guess.
mathstudent55
  • mathstudent55
Then the 5th line is just simplification inside the square brackets.
unimatix
  • unimatix
That all makes sense. It's pretty straightforward. Thank you!
mathstudent55
  • mathstudent55
Finally the last line which is on the next response above, is simply taking out a factor of -4 from inside the parentheses and bringing it out.
mathstudent55
  • mathstudent55
You're welcome.

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