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EmmaH Group Title

Factor the trinomial: 4x^2 - 8x - 5 (2x + 1)(2x - 5) (4x + 1)(x - 5) (4x - 5)(x + 1) prime

  • one year ago
  • one year ago

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  1. EmmaH Group Title
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    how do i solve this?

    • one year ago
  2. amistre64 Group Title
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    trying to pick from a multiple choice is simpler than trying to work thru the math. I always separate taking multiple choice from actually doing math.

    • one year ago
  3. amistre64 Group Title
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    we can deduce the the FL parts of foil are the same for every given option. if we can reduce this to the proper OI part that would reduce the work load

    • one year ago
  4. EmmaH Group Title
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    im not trying to find the answer im trying to understand it

    • one year ago
  5. EmmaH Group Title
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    i know that the first thing i have to do it multiply the first and last number right?

    • one year ago
  6. EmmaH Group Title
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    which would be -20

    • one year ago
  7. amistre64 Group Title
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    now multiply the "inside" numbers

    • one year ago
  8. EmmaH Group Title
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    um i thought i was suppose to find the factors of -20 then use that factors that add to -8?

    • one year ago
  9. amistre64 Group Title
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    im not sure if we are seeing the same thing at the moment. take the forms: (a+b)(c+d) if ac equals the first term; and bd equals the last term; then we need to determine the middle term by: ac+bd

    • one year ago
  10. amistre64 Group Title
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    i know its not this one, but to example my idea ... (4x - 5)(x + 1) 4x*x = 4x^2 -5*1 = -5 and our given setup is 4x^2 -8x -5 all the options give us the same first and last terms so we need to determine a suitable "middle" setup

    • one year ago
  11. EmmaH Group Title
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    im pretty sure im not doing foil im ffactoring the trinomial

    • one year ago
  12. amistre64 Group Title
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    (4x - 5)(x + 1) 4x * 1 = 4x -5 * x = -5x 4x - 5x is not equal to -8x so we can rule that out

    • one year ago
  13. EmmaH Group Title
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    ok so its -10,2?

    • one year ago
  14. amistre64 Group Title
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    factoring produces a foil; and foil results in the equation they want factored -10,2 is perfect for us

    • one year ago
  15. EmmaH Group Title
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    so id write it like thsi? 4x^2+2x-10x-5?

    • one year ago
  16. amistre64 Group Title
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    that looks suitable to me; i was never really good at the factor by grouping stuff :)

    • one year ago
  17. EmmaH Group Title
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    so then what do i do thats no9t one of te answers?

    • one year ago
  18. amistre64 Group Title
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    then you group it, and factor our the groups, then factor again ... which is why i never really go that route in these ... (4x^2+2x)+(-10x-5) 2x(2x+1)-5(2x+1) (2x+1)(2x-5)

    • one year ago
  19. rosedewittbukater Group Title
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    I responded to your other question of this.

    • one year ago
  20. rosedewittbukater Group Title
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    4x^2 - 8x - 5 Ok so I'll try to explain this the way I understand it. Usually to factor you have to find 2 numbers that multiply to equal the last number, and add up to equal the number part of the middle. I'll just use \[ax ^{2}+bx+c\] as an example equation. So you have to find 2 numbers that multiply to get c, and add up to get b. Then you would usually take those 2 numbers and use it to factor and get (x+?)(x+?). But since there is a number for a that is not 1 in your equation, it's a little different. First, multiply the first and last numbers 4 and -5 (a and c in the example equation). You get -20. So try to find 2 number that are factors of -20 and add up to equal 8. Just start to list the factors off on paper or in your head. Remember to include the negatives. -1, 20 1, -20 -2, 10 2, -10 These add up to equal -8 Then put 2 and -10 to separate the middle part of the equation. \[4x ^{2}+2x-10x-5\]Then group the equation into 2 separate parts. \[(4x ^{2}+2x)+(-10x-5)\]Then you take out any common factors from the parentheses.\[2x(2x+1)-5(2x+1)\]And then group together the parts you took out, which are the 2x and -5. same.\[(2x-5)(2x+1)\]It becomes this because 2x+1 are the There's your answer. Hope this helps!

    • one year ago
  21. EmmaH Group Title
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    thank you all soooooooooo much!!!!

    • one year ago
  22. amistre64 Group Title
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    rose has a pretty detailed answer; not alot of interaction allowed in it, but a pretty detailed answer nonetheless :)

    • one year ago
  23. GiGIgo3 Group Title
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    • one year ago
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