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anonymous
 one year ago
How to factor 2x^2+5x3???
anonymous
 one year ago
How to factor 2x^2+5x3???

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Have you learned the quadratic formula?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes. Would this be applied here?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yes the best way to factor in my opinion is to first find the roots using the quadratic formula. What roots do you get?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1ok that looks like the value of b^2  4ac

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1notice that since the discriminant is equal to 1, which is a positive perfect square, this means we can factor

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see... so how do we proceed?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I attempted to utilize the AC method when going about factoring the trinomial but ended up with an incorrect result.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1so do you have this written down? \[\Large x = \frac{b\pm\sqrt{b^24ac}}{2a}\] \[\Large x = \frac{5\pm\sqrt{5^24*2*(3)}}{2*2}\] \[\Large x = \frac{5\pm\sqrt{49}}{4}\] notice how 49 is under the root and NOT 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes; would we proceed by simplifying the division and the square root? 5+/7/4= 1/2 or 3.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1correct, the two roots are 1/2 and 3

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1if 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?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1then if x = 3, we can add 3 to both sides to get x+3 = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see where you're going now, by discovering the roots, it's much easier to find the factors.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1so the two roots x = 1/2 or x = 3 turns into these two equations 2x1 = 0 or x+3 = 0

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yes, we then use the zero product property to go from `2x1 = 0 or x+3 = 0` to `(2x1)(x+3)=0`

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the last thing to do is to expand out `(2x1)(x+3)` and see if you get `2x^2+5x3`

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0While 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
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.1you 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
 one year ago
Best ResponseYou've already chosen the best response.1you 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
 one year ago
Best ResponseYou've already chosen the best response.0I guess I would remember to do it as the factors typically need to be written with whole numbers.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah that's a good general rule to have so it's as simple as possible

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much for your thorough response!

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you're welcome
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