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anonymous

  • 5 years ago

Simplify completely: 6x^2+5x+1 ----------- 27x^3+1

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  1. anonymous
    • 5 years ago
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    I know the 27x^3+1 is a perfect cube: cuberoot 27= 3 cuberoot x^3=x cuberoot 1=1

  2. anonymous
    • 5 years ago
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    but the top can't be factored. I think I can rewrite this as (3x+1)^3 But would that make the answer 6x^2+5x+1 ----------- ? (3x+1)^3 I can't help but think there is more to this.

  3. anonymous
    • 5 years ago
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    factorize the denominator..an equation in the form a^3 + b^3 = (a+b)(a^2-ab+b^2) also factorize the numerator. the numerator can be factorized by using the formula x = \[-b \pm \sqrt{D} \div 2a\] where the quadratic is in the form ax^2 + bx + c

  4. anonymous
    • 5 years ago
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    D = b^2 - 4ac

  5. anonymous
    • 5 years ago
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    so (3x+1)(9x^2-3x+1^2)?

  6. anonymous
    • 5 years ago
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    yeah thats the denominator.

  7. anonymous
    • 5 years ago
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    Or: \[(3x+1)(9x^2-3x+1)\]

  8. anonymous
    • 5 years ago
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    yep

  9. anonymous
    • 5 years ago
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    So the numerator just stays the same then?

  10. anonymous
    • 5 years ago
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    no. factorize the numerator using the formula i gave you..the formula gives you the roots of the equation. if a is the 1st root and b is the 2nd root then the numerator can be re written as (x-a)(x-b)

  11. anonymous
    • 5 years ago
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    It won't work that way because there are 2 + signs, which means the signs must stay the same. Nothing but 1*1 will = +1

  12. anonymous
    • 5 years ago
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    Well, one sec...

  13. anonymous
    • 5 years ago
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    it will. D = 1. 25- (6*4)

  14. anonymous
    • 5 years ago
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    \[(6x-1)(x-1) = 6x^2-6x-1x+1 = 6x^2-7x+1\]

  15. anonymous
    • 5 years ago
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    That doesn't match the equation. Where did you get a 4?

  16. anonymous
    • 5 years ago
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    the formula i gave you is b^2 - 4ac. hence the 4..

  17. anonymous
    • 5 years ago
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    But you can't use negatives when both signs in the equation are positive.

  18. anonymous
    • 5 years ago
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    \[6x^2+5x+1\] That means you have to use + signs to factor from what I have been told.

  19. anonymous
    • 5 years ago
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    nope..for any quadratic expression of the form \[Ax ^{2} + Bx + C\] the formula for the roots are \[-B \pm \sqrt{D} / 2A \]

  20. anonymous
    • 5 years ago
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    What is D then? The entire equation?

  21. anonymous
    • 5 years ago
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    no..i told you D is the discriminant of the equation which is equal to B^2 - 4AC

  22. anonymous
    • 5 years ago
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    So \[-5 \pm \sqrt{5^2-4(6)(1)} \div 2(6)\] ?

  23. anonymous
    • 5 years ago
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    exactly..

  24. anonymous
    • 5 years ago
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    \[-5 \pm \sqrt{25-24} \div 12\]

  25. anonymous
    • 5 years ago
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    \[-5 \pm 1 \div 12\]

  26. anonymous
    • 5 years ago
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    -6 ---- 12 or -4 ---- 12

  27. anonymous
    • 5 years ago
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    -2 --- 4 -1 --- 12

  28. anonymous
    • 5 years ago
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    But they are fractions. How am I supposed to plug those into the equation?

  29. anonymous
    • 5 years ago
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    -1/3 and -1/2

  30. anonymous
    • 5 years ago
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    the equation can be written as (x+1/3)(x+1/2) = 0.

  31. anonymous
    • 5 years ago
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    Yes, I typed them wrong, I see that

  32. anonymous
    • 5 years ago
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    so -1/2 or -1/3

  33. anonymous
    • 5 years ago
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    But how does it help if it equals 0? That would solve for x for the top equation, but I am simplifying.

  34. anonymous
    • 5 years ago
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    I thought the quadratic formula was for determining the values of x?

  35. anonymous
    • 5 years ago
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    sorry it doesnt equal 0

  36. anonymous
    • 5 years ago
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    its just that the numerator can be re written as (x+1/3)(x+1/2)

  37. anonymous
    • 5 years ago
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    And also, (x+1/3_)x+1/2) \[\neq\] to \[6x^2+5x+1\]

  38. anonymous
    • 5 years ago
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    Sorry, *(x+1/3)(x+1/2) \[\neq 6x^2+5x+1\]

  39. anonymous
    • 5 years ago
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    Thats why I was saying it can't be factored.

  40. anonymous
    • 5 years ago
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    it is equal to that divided by 6. you can always multiply and divide 6 and retain the eqn.

  41. anonymous
    • 5 years ago
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    But it is already in a fraction. I can't have a fraction over a fraction and call it simplified.

  42. anonymous
    • 5 years ago
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    Okay, figured something out, but didn't need quadratic formula for it.

  43. anonymous
    • 5 years ago
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    \[\frac{(1+2 x)}{(1+6 x)^2} \]

  44. anonymous
    • 5 years ago
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    Sorry. The denominator should be: \[\left(1-3 x+9 x^2\right) \] \[\frac{6 x^2+5 x+1}{27 x^3+1}=\frac{1+2 x}{1-3 x+9 x^2} \]

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