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Jravenv

  • one year ago

Help please best answer rewarded screencap in comments

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  1. Jravenv
    • one year ago
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  2. ybarrap
    • one year ago
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    Break this up into pieces Can you factor \(m^2-4\)?

  3. Jravenv
    • one year ago
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    Nope I have no idea how to do any of this

  4. ybarrap
    • one year ago
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    $$ (m+2)(m-2)=m^2-2m+2m-4=m^2-4 $$ Do you see that?

  5. UsukiDoll
    • one year ago
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    we can factor out like terms and then cancel anything that's the same in the numerator and the denominator.

  6. UsukiDoll
    • one year ago
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    oh hi

  7. UsukiDoll
    • one year ago
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    \[\frac{3m-6}{4m+12}\] let's consider this part of the problem... there is a number in common on 3m -6. What number can I factor out? also what is 3/3 and -6/3 similarly what number can I factor out? what is 4/4 and 12/4 for the denominator..

  8. UsukiDoll
    • one year ago
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    I can pull out a 3 for 3m-6 and I will have 3(m-2). I can pull out a 4 for 4m+12 and I will have 4(m+3)

  9. UsukiDoll
    • one year ago
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    \[\frac{m^2+5m+6}{m^2-4}\] this requires factoring as well. We have a perfect square on the bottom of the denominator. now we just need to factor the numerator which is \[m^2+5m+6\] We need to focus on all combinations that make 6 and use addition or subtraction to make a 5 the combinations of 6 are 6 and 1 1 and 6 2 and 3 3 and 2

  10. UsukiDoll
    • one year ago
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    6 and 1 isn't going to work . If I add them together it's 7... subtract and it's a 5. Well why can't we use this? Because our equation has double + signs... so it requires all numbers to be positive.

  11. UsukiDoll
    • one year ago
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    2 and 3 will work just fine. Not only is it 2 x 3 = 6, but 2+3 =5 so our \[m^2+5m+6 \rightarrow (m+2)(m+3)\]

  12. UsukiDoll
    • one year ago
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    \[\frac{3(m-2)}{4(m+3)} \times \frac{(m+2)(m+3)}{(m+2)(m-2)}\]

  13. UsukiDoll
    • one year ago
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    there are a bunch of terms to cancel out I see (m+3) on the numerator and denominator...the same case happens for (m-2) and (m+2). ANd then you are left with \[\frac{3}{4}\]

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spraguer (Moderator)
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