## anonymous one year ago How to factor 2x^2+5x-3???

1. jim_thompson5910

Have you learned the quadratic formula?

2. anonymous

Yes. Would this be applied here?

3. anonymous

@jim_thompson5910

4. 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?

5. anonymous

25-(24)=1

6. jim_thompson5910

ok that looks like the value of b^2 - 4ac

7. jim_thompson5910

notice that since the discriminant is equal to 1, which is a positive perfect square, this means we can factor

8. anonymous

I see... so how do we proceed?

9. anonymous

I attempted to utilize the AC method when going about factoring the trinomial but ended up with an incorrect result.

10. 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

11. anonymous

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

12. jim_thompson5910

correct, the two roots are 1/2 and -3

13. 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?

14. anonymous

Certainly.

15. jim_thompson5910

then if x = -3, we can add 3 to both sides to get x+3 = 0

16. anonymous

I see where you're going now, by discovering the roots, it's much easier to find the factors.

17. jim_thompson5910

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

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

the last thing to do is to expand out (2x-1)(x+3) and see if you get 2x^2+5x-3

20. 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?

21. 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

22. 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

23. 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

24. anonymous

I guess I would remember to do it as the factors typically need to be written with whole numbers.

25. jim_thompson5910

yeah that's a good general rule to have so it's as simple as possible

26. anonymous

Thank you so much for your thorough response!

27. jim_thompson5910

you're welcome