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Some factorizing and other equations, please help

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data
    1, 2. Factorize completely:
    80t^5-5t

    3X-9X^2+12

    solve: 3. 1/2x = 1/6+1/3X

    4. A right-angle triangle has the hypotenuse 16.0, one of the short sides is 3 cm longer than the other. Area and perimeter?
    2. Relevant equations



    3. The attempt at a solution
    1. 5t(16t^4-1)
    2. x(3-9x)+12 are they completely factorized though?
    3. I cross multiplied until I got: 18X = 5X^2 + 12X... can't get any further
    4. no success
     
    Last edited: Feb 26, 2008
  2. jcsd
  3. Feb 26, 2008 #2

    HallsofIvy

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    Notice that 16t2= (4t)2 and 1= 12. There is a special formula for a "difference of two squares".

    That isn't "factored" at all! It is not ( )( ). You should be able to see immediately that you can factor out a 3: 3x2- 9x+ 12= 3(x2- 3x+ 4). Now can you factor 4 into two integers whose "sum" is -3?

    You mean, I think, "solve 1/(2x)= 1/6+ 1/(3x). (What you wrote could be interpreted as (1/2)x= 1/6+ (1/3)x.) You didn't really need to multiply the left side by 18: 6 would have done it. 1/6+ 1/(3x)= 1/6+ 2/(6x)= (x+ 2)/6x Any way, "cross multiplying": multiply the left by 18x and the right by 2x gives 18x= 2x2+ 4x. I find it confusing to "cross multiply". I would prefer to say "multiply both sides by the least common denominator" which is 6x: 6x(1/(2x))= 6x(1/6)+ (6x)(1/3x) so 3= x+ 2 (not what I got above- I messed up the "cross multiply"!) . Can you solve 3= x+ 2?

    A right-angle triangle has the hypotenuse 16.0, one of the short sides is 3 cm longer than the other. Area and perimeter?
    Well, a "relevant equation" would be the Pythagorean theorem. If we call one side "x" and the other side is "3 cm longer", what would the other side be? Now put those into the Pythagorean theorem to get the equation you need to solve. Once you know the lengths of the sides, it is easy to find the area and perimeter of the triangle.
     
  4. Feb 26, 2008 #3
    I've solved the last one. But the two first I can't get at all :(....
     
  5. Feb 26, 2008 #4
    The first one: 5t(16t4-1) can be further factorized by a great amount. And remember:
    (a - b)(a + b) = a2 - ab + ab - b2 = a2 - b2

    For the next one: (x + a)(x +b) = x2 + bx + ax + ab = x2 + x(a + b) + ab

    Try to compare your equations with these.
     
  6. Feb 27, 2008 #5

    HallsofIvy

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    What are the possible ways of factoring 4 into two integers? Do any of those add up to -3?
     
  7. Feb 27, 2008 #6
    For the second question, the original equation is -9x2 + 3x + 12, not 3x2 - 9x + 12, but HallsofIvy's suggestions for factoring still apply.
     
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