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anonymous

  • one year ago

Hey guys i have three questions can you help me? Ive just started college and I need help with this basic algebra. 1. 2a^2+7ab+3b^2 2. 3a^2-8ab-28b^2 3. 36x^2-12x+1 if you could also explain to me how to do this that would be even better.

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  1. anonymous
    • one year ago
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    please help me!!

  2. Nnesha
    • one year ago
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    quadratic equations: there are 4 method to factor out the quadratic equations

  3. Nnesha
    • one year ago
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    s*

  4. Nnesha
    • one year ago
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    i'll go with the easy one Ac method also known as headphone method where you have to find two numbers when you multiply them you should get product of AC and when you add them yo should get middle term

  5. anonymous
    • one year ago
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    for the first one I would start it as (2a+ ?b) (a+?b)

  6. anonymous
    • one year ago
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    so i would multiply the 2 and the 3 to get the middle term?

  7. Nnesha
    • one year ago
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    yes right: find two numbers when you multiply them you should get 6 but when you add them you should get 7ab

  8. anonymous
    • one year ago
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    that would be 2a^2+8ab+6b^2 thats not right

  9. Nnesha
    • one year ago
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    hmm

  10. anonymous
    • one year ago
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    okay im not following why do i add?

  11. anonymous
    • one year ago
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    or would this one be prime?

  12. Nnesha
    • one year ago
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    you should find `two` numbers when you `add` them you should get `middle` term but when you multiply same numbers you should get product of ac what two numbers would you multiply to get 6 but when you add them you get 7 ?

  13. anonymous
    • one year ago
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    1 and 6

  14. Nnesha
    • one year ago
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    yes right leading coefficient isn't so we have to use factor by groping method |dw:1440352936888:dw| carry down the first and last term and write the numbers u just found in the middle

  15. Nnesha
    • one year ago
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    |dw:1440353069776:dw| make a group of `2` terms find the common factor of `1st two terms and then common factor of last 2 terms

  16. Nnesha
    • one year ago
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    \[\huge\rm \color{Red}{2a^2+6ab}+\color{blue}{1ab+3b^2}\] what is common in first two terms ?

  17. anonymous
    • one year ago
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    so that makes (a+3b) (2a+b) right?

  18. anonymous
    • one year ago
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    because you factor out the highest factor

  19. anonymous
    • one year ago
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    of each group

  20. Nnesha
    • one year ago
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    yes that's correct!

  21. anonymous
    • one year ago
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    okay cool

  22. Nnesha
    • one year ago
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    you can foil it to check your answer (2a+b)(a+3b)=2a^2+7ab+3b^2

  23. anonymous
    • one year ago
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    okay i foiled it and i got the same answer so its correct let me try the other ab problem the problem 36x^2-12x+1 I am also still having troubles with

  24. Nnesha
    • one year ago
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    do you want to try it first ? :)

  25. anonymous
    • one year ago
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    im trying the other ab one first.

  26. Nnesha
    • one year ago
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    okay :=)

  27. anonymous
    • one year ago
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    okay im stuck i have 3a^2 -8ab-28b^2 I multiplied the 3 by the 28 and got 84 then used the factors 6 and 14 then I got 3a^2-6ab-42ab-28b^2 and then i put them in groups like you said (3a^2-6ab) (-42ab -28b^2) then i took 3 and a out of the first group and negative 14b out of the second group then I was left with (3a-14b) (a-2b) (3a+2b) im confused on what i did wrong

  28. Nnesha
    • one year ago
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    oky so by just looking at two numbers i can say thats wrong 6 times 28 =64 okay you have to be careful about signs

  29. anonymous
    • one year ago
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    okay

  30. Nnesha
    • one year ago
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    \[ \huge\rm 3a^2 \color{Red}{-}8ab\color{Red}{-}28b^2 \] when you multiply two numbers you should get negative 84 bec `3 times -28=-84` and when you add them you should get `-8`

  31. Nnesha
    • one year ago
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    so 6 times 14 = 84 not negative but what if take -14 times 6 =

  32. anonymous
    • one year ago
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    that equals the -8 that is needed

  33. Nnesha
    • one year ago
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    yes right thats what we need and thats two number you should put in the middle http://prntscr.com/880qre

  34. Nnesha
    • one year ago
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    \[\rm 3a^2+6ab-14ab+28b^2\]

  35. anonymous
    • one year ago
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    okay like this 3a^2 -14ab+6ab-28b^2

  36. anonymous
    • one year ago
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    then i separate them into groups and factor

  37. Nnesha
    • one year ago
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    yes right! :+)

  38. anonymous
    • one year ago
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    okay i got the correct answer so then it is (3a-14b)(a+2b) and i used foil to make sure it was correct and got 3a^2-8ab-28b^2

  39. Nnesha
    • one year ago
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    yes right

  40. anonymous
    • one year ago
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    okay now on to the last problem this one i have started it as |dw:1440355529657:dw|

  41. anonymous
    • one year ago
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    wait hold on i think i just figured it out

  42. Nnesha
    • one year ago
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    how do you know it would be one minus and one plus parentheses ? i mean \[\huge\rm (6x-?)(x+??)\]

  43. anonymous
    • one year ago
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    oh wow i feel dumb hahha i didnt realize that the number was supposed to be 1 so i did this |dw:1440355938178:dw|

  44. anonymous
    • one year ago
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    well thank you natasha

  45. Nnesha
    • one year ago
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    |dw:1440356037300:dw| that's right !

  46. Nnesha
    • one year ago
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    there is ano|dw:1440356070415:dw| 0ther way of factoring quadratic i would like to show that one too ~find factor of first and last term then cross multiply if you get middle term then the numbers you multiplid are factors of the quadratic equation

  47. Nnesha
    • one year ago
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    |dw:1440356266572:dw|