kdarnell08 2 years ago How about 4x^2 + 81- 36x

1. funinabox

what exactly is the question

2. kdarnell08

i dont know how to do the problem, and i want someone to walk me throught the steps so i can understand the futture questions. 4x^2+81-36x

3. Hero

When factoring any quadratic, the goal is always the same. Find two numbers (x and y) that add to get b, but multiply to get ac. x + y = b xy = ac

4. Hero

In this case, we need to find: x + y = -36 xy = 324

5. Hero

The only two numbers that work is x = -18 y = -18

6. Hero

Anytime both x and y are the same, it means you will have a perfect square.

7. Hero

4x^2 - 36x + 81 4x^2 - 18x - 18x + 81 2x(2x - 9) - 9(2x - 9) (2x - 9)(2x - 9) (2x - 9)^2

8. kdarnell08

alright. so how about 4y^2 +16y+16 How would that work..

9. Hero

It works the same way: x + y = 16 xy = 64 Find x and y

10. Hero

I'm pretty sure you can find the two numbers

11. Hero

It doesn't require doing any actual Algebra. All you do is ask yourself, "What two numbers add to get 16, yet multiply to get 64?".

12. Hero

In this case, the two numbers are the same.

13. Hero

I'm not going to your homework for you. I'll show you how to do one problem. But, don't think I'm going to show you how to do every single problem. You need to figure this one out. The steps are exactly the same as the previous problem.

14. kdarnell08

i dont expect you to do it for me lol. i actually need to learn it. im not in hs like a bunch of the people on there. i am in nursing school and personally do not want to have to repay for the class.

15. Hero

Okay, so did you find the two numbers yet?

16. funinabox

if you are having trouble, a simple way of solving this is substitution. x + y = 16 xy = 64 or in other words y = 16 - x x(16 - x) = 64 further simplified... 16x - x^2 = 64 x^2 - 16x + 64 = 0 then you can factor and solve

17. funinabox

which is a much more simple polynomial to answer than the previous one

18. Hero

Funny thing is, she would still need to find the same two numbers either way.

19. funinabox

i know i realized that after i posted it :D but it's still a more simple polynomial to look at, at least in terms of co-efficients

20. kdarnell08

ok. i think i got it.

21. kdarnell08

4y^2 + 16y +16 4(y^2 + 4y + 4) 4(y^2+ 2y + 2y+ 2) 4 [y(y+2) + 2(y+2)] =4(y+2)^2

22. kdarnell08

right? yes no? lol ive always been terrible with numbers :/

23. Hero

Good job

24. kdarnell08

sheesh. glad i dont ddeal with this in my profession ha

25. Hero

You have to think of math as "playing with numbers", rather than "a chore".

26. kdarnell08

its just too difficult for me to understand. anyting else you can put in front of me and i get it right away, and sure certain parts of math i can too, but when it comes to this i just...ugggg. wanna curl up in a ball and cry

27. Hero

Most of the time, students that don't get it are usually those who are not very good with seeing the relationships between addition and multiplication. Either that or they don't have their multiplication tables well memorized.

28. Hero

This concept is definitely more "mental math" compared to other algebra concepts.

29. kdarnell08

my tables are fine...and maybe i do have a problem with my adding and multiplying and i just dont realize it? cuz ive never gotten the factoring polynomials things. not now, not in hs

30. Hero

I'm sorry to hear that.

31. kdarnell08

hey. only gotta deal with this until march so i think i can stick it out

32. Hero

If you have difficulty with "finding the two numbers that add to get one number, but multiply to get another", then your problem is not with factoring, but with the multiplication/addition relationships.

33. kdarnell08

yeah. thats what i thought.

34. kdarnell08

well thank you! i appreciate it very much!! back to other homework. Happy Holidays!