1. anonymous

@jagr2713

2. myininaya

hint: this may help" $x^2+bx+\frac{c}{a} \\ \frac{1}{a}(ax^2+bax+\frac{c(a)}{a}) \\ \frac{1}{a}(ax^2+bax+c)$

3. anonymous

ok so my equation should look like $1/4(4x^2+3(4)x+9)$

4. myininaya

yep it is like dividing everything by 4 at the same time as multiplying everything by 4 and we know 4/4=1 so yep

5. myininaya

and 3(4) can be written as 12 of course

6. myininaya

you can also see if you need to factor anymore

7. anonymous

right, so now my equation should be $x^2+12+9$

8. myininaya

well you are missing 1/4( ) and a 4 in front of x^2 and a x next to the 12

9. myininaya

$\frac{1}{4}(4x^2+12x+9)$

10. myininaya

anyways let's look at the thing in ( )

11. myininaya

can you find two numbers that multiply to be 4(9)=36 and add up to be 12

12. anonymous

something that are multiples to 36 and add to 12?

13. myininaya

yep

14. anonymous

6 and 6

15. myininaya

yes yes! :)

16. myininaya

$\frac{1}{4}(4x^2+6x+6x+9)$ now we can ignore the 1/4 and bring it down later but we need to factor the thing inside the ( )by grouping

17. myininaya

we will bring the 1/4 down later $4x^2+6x+6x+9 \\ (4x^2+6x)+(6x+9) \\ 2x(2x+3)+(6x+9)$ you try factoring the 6x+9

18. anonymous

(6x+9) will equal 3x(2x+3)

19. myininaya

well 9 doesn't have a factor of x so you cannot factor an x out of it unless you divide by x but that is making is nastier than it should be 6x+9=3(2x+3)

20. myininaya

we will bring down 1/4 later $2x(2x+3)+3(2x+3)$ now one last step

21. anonymous

oh ok, that makes sense!

22. anonymous

I'm not sure how to do the last step

23. myininaya

example to factor a(b+c)+m(b+c) notice both terms both a(b+c) and m(b+c) have the common factor (b+c) so I can factor (b+c) out of both terms in the sum (b+c)(a+m)

24. myininaya

so you see (2x+3) in both terms the terms being 2x(2x+3) and 3(2x+3)

25. myininaya

what is left of the first term when factoring out the (2x+3) is 2x and of the second term is +3

26. myininaya

2x(2x+3)+3(2x+3) (2x+3)(2x+3)

27. myininaya

don't forget to bring down the 1/4

28. anonymous

so its just 1/4(2x+3)(2x+3)

29. myininaya

or 1/4(2x+3)^2

30. myininaya

or...

31. myininaya

$\frac{1}{4}(2x+3)^2 \\ \frac{1}{2^2}(2x+3)^2 \\ \frac{1^2}{2^2}(2x+3)^2 \\ (\frac{1}{2}(2x+3))^2 \\ (\frac{2x+3}{2})^2$ there is another or :p

32. myininaya

you could separate the fraction inside the ( )^2

33. myininaya

but anyways yeah you are done

34. anonymous

so its just $(x+\frac{ 3 }{ 2 })^2$

35. myininaya

yep

36. anonymous

Thank you so much, I really appreciate your help!

37. myininaya

if we would have noticed the following to begin with: $x^2+3x+(\frac{3}{2})^2=(x+\frac{3}{2})^2$ we wouldn't had so much work

38. myininaya

$x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2$

39. anonymous

oh yes! That's very true, we missed that. Oh well :)

40. myininaya

for example: $x^2+5x+(\frac{5}{2})^2=(x+\frac{5}{2})^2 \\ x^2-5x+(\frac{5}{2})^2=(x-\frac{5}{2})^2 \\ x^2-3x+(\frac{-3}{2})^2=(x-\frac{3}{2})^2 \\ x^2-2x+(\frac{-2}{2})^2=(x-1)^2$

41. anonymous

that actually makes perfect sense, thank you so much for showing me that formula! It's a lot easier than all the other ways I have been taught.

42. myininaya

no problem :)