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calculusxy

  • one year ago

factoring quadratic expressions

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  1. calculusxy
    • one year ago
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    \[18n^2 - 8\]

  2. calculusxy
    • one year ago
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    @phi

  3. Nnesha
    • one year ago
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    take out the common factor what is GCF(greatest common factor ) ?

  4. calculusxy
    • one year ago
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    2?

  5. Nnesha
    • one year ago
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    right take out 2 from 18n^2 -8 or in other words divide both terms by common factor \[2(\frac{ 18n^2 }{ 2 }-\frac{ 8 }{ 2 })\]

  6. calculusxy
    • one year ago
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    2(9n^2 - 4)

  7. phi
    • one year ago
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    you should always be on the look out for "difference of squares"

  8. calculusxy
    • one year ago
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    @phi I don't understand...

  9. Nnesha
    • one year ago
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    9 and 4 are perfect square roots and the negative between both terms so you can apply the difference of squares \[\huge\rm a^2-b^2 =(a-b)(a+b)\]

  10. calculusxy
    • one year ago
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    so how would i implement it on 9^2 - 4?

  11. Nnesha
    • one year ago
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    this is how i do it take square roots of both terms write in two parentheses (sqrt of 1st term `+` sqrt of 2nd term) (sqrt of 1st term `-` sqrt of 2nd term ) lolol

  12. calculusxy
    • one year ago
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    okay.. so would it be (3+4)(3-4)?

  13. Nnesha
    • one year ago
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    what about n it's 9n^2 - 4

  14. calculusxy
    • one year ago
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    (3n+4)(3n-4)

  15. Nnesha
    • one year ago
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    looks good don't forget the common factor 2(3n+4)(3n-4)

  16. Nnesha
    • one year ago
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    ohh wait no

  17. Nnesha
    • one year ago
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    |dw:1444163520807:dw| take square root of both termz

  18. Zarkon
    • one year ago
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    \[9n^2 - 4 =(3n)^2-2^2\]

  19. calculusxy
    • one year ago
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    sorry but this is not clear to me at all

  20. calculusxy
    • one year ago
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    i need more help on this

  21. Nnesha
    • one year ago
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    alright let's start when the coefficient of both terms are perfect squares and the exponent of the variable is even and the negative sign between the terms then you can use difference of squares hint

  22. anonymous
    • one year ago
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    im not that good at math. I would try to help you but I would end up failing.

  23. anonymous
    • one year ago
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    no thats not true you wont fail.

  24. anonymous
    • one year ago
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    and if you do like there is a saying First comes frailer second comes success.

  25. Nnesha
    • one year ago
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    \[\huge\rm x^2 -4\] even power and both coefficient are perfect squares since the negative sign between them we can use the hint \[\huge\rm\color{ReD}{ a}^2 -\color{blue}{b}^2 = (a-b)(a+b)\] we have to rewrite 4 as 2 exponent so 2^2 = 4 \[\huge\rm \color{ReD}{x}^2-\color{blue}{2}^2=(\color{Red}{x}-\color{blue}{2}(\color{Red}{x}+\color{blue}{2}) \]

  26. anonymous
    • one year ago
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    My god.

  27. anonymous
    • one year ago
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    that wuill take a while. lol.

  28. Nnesha
    • one year ago
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    that's what zarkon mentioned above that's same as taking square roots of both terms

  29. anonymous
    • one year ago
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    ahh

  30. Nnesha
    • one year ago
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    |dw:1444164108305:dw| one with negative sign and one with positive

  31. Nnesha
    • one year ago
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    now make sense or no ?? hm

  32. calculusxy
    • one year ago
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    u kinda do.. but i m still trying to understand it

  33. Nnesha
    • one year ago
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    alright take ur time let me know what u don't understand :=) will try to explain

  34. calculusxy
    • one year ago
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    if we have 16^2 - 49, would we have (4 - 7)(4+7) or something else?

  35. Nnesha
    • one year ago
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    there should be a variable

  36. calculusxy
    • one year ago
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    where?

  37. Nnesha
    • one year ago
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    it should be 16x^2 -49

  38. Nnesha
    • one year ago
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    remember if it's 16^2-49 both are `like` terms so we can just combine them

  39. calculusxy
    • one year ago
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    oh okay... so would we have that as (4n + 7)(4n - 7) ?

  40. Nnesha
    • one year ago
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    looks good

  41. calculusxy
    • one year ago
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    so for my original problem the answer would be:

  42. calculusxy
    • one year ago
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    \[2(9n^2 - 4)\] \[9n^2 - 4 \implies (3n + 2)(3n-2)\] \[\large Answer\] \[2(3n+2)(3n-2)\]

  43. Nnesha
    • one year ago
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    \(\huge\color{Green}{\checkmark}\)

  44. Nnesha
    • one year ago
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    that's right!

  45. calculusxy
    • one year ago
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    Thank you so much! I just want to review the steps...

  46. Nnesha
    • one year ago
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    just do some example u will get the concept :=)

  47. calculusxy
    • one year ago
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    1. Find the GCF of the terms 2. Put the GCF outside of the parenthesis. 3. Divide the terms of the question by the GCF. 4. (if there are perfect squares, then we use a^2 - b^2 = (a + b)(a - b)

  48. calculusxy
    • one year ago
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    5. and then put the GCF outside

  49. Nnesha
    • one year ago
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    right for perfect squares: sign should be negative between terms both numbers should be perfect squares even power!

  50. calculusxy
    • one year ago
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    i have a small question... if we have -x^2, can we turn the x to be a positive ?

  51. calculusxy
    • one year ago
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    ...since the exponent is a positive, i thought that that's possible

  52. calculusxy
    • one year ago
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    @Nnesha

  53. Nnesha
    • one year ago
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    it depends on 2nd term let's say we have \[\huge\rm -x^2 - 9\] we can factor out the negative one \[\large\rm -1(x^2+9)\] now there is a positive sign we can't factor it out

  54. Nnesha
    • one year ago
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    and if we have \[\huge\rm -x^2+9\] take out the negative one a should be positive

  55. calculusxy
    • one year ago
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    -x^2 + y^3... i think this was a question on a test that i took

  56. calculusxy
    • one year ago
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    like we had to divide with exponents and that was in the denominator... wait. let me write it out \[\frac{ 1 }{ -x^{-2}y^3 }\]

  57. calculusxy
    • one year ago
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    i moved -x^{-2} up to the numerator. \[\frac{ -x^2 }{ y^3 } \implies \frac{ x^2 }{ y^3}\]

  58. calculusxy
    • one year ago
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    would that be valid?

  59. Nnesha
    • one year ago
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    ohhh no wait why did you flip the fraction ?

  60. Nnesha
    • one year ago
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    IF IF `exponent ` is negative then you should flip the fraction

  61. calculusxy
    • one year ago
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    i only moved -x^{-2} since the exponent was negative

  62. Nnesha
    • one year ago
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    wait a sec its lagging let me refresh the page

  63. Nnesha
    • one year ago
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    \[\huge\rm \frac{ 1 }{ -x^2y^3}\] is this ur question ?

  64. calculusxy
    • one year ago
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    no\[\huge \frac{ 1 }{ -x^{-2}y^3 }\]

  65. Nnesha
    • one year ago
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    ohh okay make sense

  66. Nnesha
    • one year ago
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    right \[\huge\rm \frac{ 1 }{ -x^{-2}y^3 } =- \frac{ x^2 }{ y^3 }\] sign would be at the top never leave the negative sign in the denominator negative times positive = negative

  67. Nnesha
    • one year ago
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    you forgot the negative sign variables are correct :=)

  68. calculusxy
    • one year ago
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    ok thank you!

  69. Nnesha
    • one year ago
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    np :=) good luck!

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