anonymous
  • anonymous
How to factor 2x^2+5x-3???
Mathematics
jamiebookeater
  • jamiebookeater
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jim_thompson5910
  • jim_thompson5910
Have you learned the quadratic formula?
anonymous
  • anonymous
Yes. Would this be applied here?
anonymous
  • anonymous

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jim_thompson5910
  • jim_thompson5910
yes the best way to factor in my opinion is to first find the roots using the quadratic formula. What roots do you get?
anonymous
  • anonymous
25-(24)=1
jim_thompson5910
  • jim_thompson5910
ok that looks like the value of b^2 - 4ac
jim_thompson5910
  • jim_thompson5910
notice that since the discriminant is equal to 1, which is a positive perfect square, this means we can factor
anonymous
  • anonymous
I see... so how do we proceed?
anonymous
  • anonymous
I attempted to utilize the AC method when going about factoring the trinomial but ended up with an incorrect result.
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
Yes; would we proceed by simplifying the division and the square root? -5+/-7/4= 1/2 or -3.
jim_thompson5910
  • jim_thompson5910
correct, the two roots are 1/2 and -3
jim_thompson5910
  • jim_thompson5910
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?
anonymous
  • anonymous
Certainly.
jim_thompson5910
  • jim_thompson5910
then if x = -3, we can add 3 to both sides to get x+3 = 0
anonymous
  • anonymous
I see where you're going now, by discovering the roots, it's much easier to find the factors.
jim_thompson5910
  • jim_thompson5910
so the two roots x = 1/2 or x = -3 turns into these two equations 2x-1 = 0 or x+3 = 0
jim_thompson5910
  • jim_thompson5910
yes, we then use the zero product property to go from `2x-1 = 0 or x+3 = 0` to `(2x-1)(x+3)=0`
jim_thompson5910
  • jim_thompson5910
the last thing to do is to expand out `(2x-1)(x+3)` and see if you get `2x^2+5x-3`
anonymous
  • anonymous
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?
jim_thompson5910
  • jim_thompson5910
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`
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
I guess I would remember to do it as the factors typically need to be written with whole numbers.
jim_thompson5910
  • jim_thompson5910
yeah that's a good general rule to have so it's as simple as possible
anonymous
  • anonymous
Thank you so much for your thorough response!
jim_thompson5910
  • jim_thompson5910
you're welcome

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