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anonymous

  • one year ago

How to factor 2x^2+5x-3???

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  1. jim_thompson5910
    • one year ago
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    Have you learned the quadratic formula?

  2. anonymous
    • one year ago
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    Yes. Would this be applied here?

  3. anonymous
    • one year ago
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    @jim_thompson5910

  4. jim_thompson5910
    • one year ago
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    yes the best way to factor in my opinion is to first find the roots using the quadratic formula. What roots do you get?

  5. anonymous
    • one year ago
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    25-(24)=1

  6. jim_thompson5910
    • one year ago
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    ok that looks like the value of b^2 - 4ac

  7. jim_thompson5910
    • one year ago
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    notice that since the discriminant is equal to 1, which is a positive perfect square, this means we can factor

  8. anonymous
    • one year ago
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    I see... so how do we proceed?

  9. anonymous
    • one year ago
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    I attempted to utilize the AC method when going about factoring the trinomial but ended up with an incorrect result.

  10. jim_thompson5910
    • one year ago
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    so do you have this written down? \[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\] \[\Large x = \frac{-5\pm\sqrt{5^2-4*2*(-3)}}{2*2}\] \[\Large x = \frac{-5\pm\sqrt{49}}{4}\] notice how 49 is under the root and NOT 1

  11. anonymous
    • one year ago
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    Yes; would we proceed by simplifying the division and the square root? -5+/-7/4= 1/2 or -3.

  12. jim_thompson5910
    • one year ago
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    correct, the two roots are 1/2 and -3

  13. jim_thompson5910
    • one year ago
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    if x = 1/2, then we can multiply both sides by 2 to get 2x = 1 then subtract the 1 from both sides to get 2x - 1 = 0 agreed?

  14. anonymous
    • one year ago
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    Certainly.

  15. jim_thompson5910
    • one year ago
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    then if x = -3, we can add 3 to both sides to get x+3 = 0

  16. anonymous
    • one year ago
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    I see where you're going now, by discovering the roots, it's much easier to find the factors.

  17. jim_thompson5910
    • one year ago
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    so the two roots x = 1/2 or x = -3 turns into these two equations 2x-1 = 0 or x+3 = 0

  18. jim_thompson5910
    • one year ago
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    yes, we then use the zero product property to go from `2x-1 = 0 or x+3 = 0` to `(2x-1)(x+3)=0`

  19. jim_thompson5910
    • one year ago
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    the last thing to do is to expand out `(2x-1)(x+3)` and see if you get `2x^2+5x-3`

  20. anonymous
    • one year ago
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    While this certainly is an easier method, I don't know if in application I will be able to know to multiply the root, x=1/2, by 2 on both sides. What tells us we need to do this?

  21. jim_thompson5910
    • one year ago
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    the goal is to get each root to turn into an equation where the right side is 0 so we can use the fact that `A = 0 or B = 0` turns into `A*B = 0`

  22. jim_thompson5910
    • one year ago
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    you dont' have to multiply by 2, but I don't like fractions which is why I try to get rid of them as soon as possible

  23. jim_thompson5910
    • one year ago
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    you could easily have x - (1/2) = 0 and x+3 = 0 turn into (x - (1/2))*(x+3) = 0 but that's a bit more complicated than it needs to be

  24. anonymous
    • one year ago
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    I guess I would remember to do it as the factors typically need to be written with whole numbers.

  25. jim_thompson5910
    • one year ago
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    yeah that's a good general rule to have so it's as simple as possible

  26. anonymous
    • one year ago
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    Thank you so much for your thorough response!

  27. jim_thompson5910
    • one year ago
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    you're welcome

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