## anonymous 5 years ago Factor Completely: 8x^3+12x^2-10x-15

1. anonymous

Everything after the $8x^3$ I could do now, but would you factor the above like so? (2x)^3

2. anonymous

$(2x)^3$ rather

3. anonymous

I tried going the right side by itself, but no divisors of 180 = -10. The closest was 10*-8

4. anonymous

I think you're going to have to factor into three linear equations. ie, (2x-a)(2x-b)(2x-c)

5. anonymous

Yech, I will try that

6. anonymous

That left me with $2x(4x^2+6x-5)-15$

7. anonymous

Trying to solve that too

8. anonymous

But the 4x^2+6-5 won't factor either.

9. anonymous

I'm sorry, 4x^2+6x-5 I multiplied 5*4 and got 20. I put in the negative sign and got -20. The only things that go into that are 10*2 and 5*4

10. anonymous

Neither of those, no matter how you switch them around, equals +6

11. anonymous

Unless the simplest factored form of the top problem is: $(2x-15)(4x^2+6x-5)$

12. anonymous

And that can't be right either. I just started long multiplication, and I already got -75 and that isn't 15

13. anonymous

Yeah, that's definitely not right.

14. anonymous

Maybe you just can't factor it and it is a trick question?

15. anonymous

I think you can, I just don't remember how to do 3rd power equations. Still plugging and chugging. ;)

16. anonymous

He does give us equations that have no solutions all of the time, among ones that look like they don't have solutions but they really do. Ugh, I am going to screw this final up so bad :P

17. anonymous

I found something on the net saying this is a cubic function and not a quadratic function

18. anonymous

That's annoying. Have you had other problems that factor out to three linear equations? ic, (ax+d)(bx+e)(cx+f)?

19. anonymous

right, that is true. Quadratic should have a highest power of x^2.

20. anonymous

Never heard of it before. But the problem I looked up had something similar and the answered turned into: $\pm i \sqrt{?}$

21. anonymous

a 3 was supposed to be in the square root box

22. anonymous

nvm, that was solving for x.

23. anonymous

$i=\sqrt{-1}$ And I doubt that is what you are working with!

24. anonymous

No matter how I break up the problem, nothing factors anywhere other than: $2x(4x^2+6x-5)-15$

25. anonymous

The next 2 problems after this are easy, so it makes me feel like I should know how to do this.

26. anonymous

look 3/4 of the page down here: http://salinesports.org/mr_frederick/Algebra%202A/Unit%206/Notes_0605.pdf

27. anonymous

I'm not very familiar with that method, but it does work.

28. anonymous

I found a formula, one sec

29. anonymous

$x=y-(b)/(3a)$

30. anonymous

I have an answer for you.

31. anonymous

The b/3a part is a fraction

32. anonymous

what is y? Look at that link if you can. I'll type out the equation here if you want/rather.

33. anonymous

I will let you go on this, I have bothered you for way too long :P

34. anonymous

It is calling it a depressed cubic

35. anonymous

36. anonymous

That is what it says to do

37. anonymous

Our objective is to find a real root of the cubic equation ax^3+bx^2+cx+d=0.

38. anonymous

is what it says

39. anonymous

But anyway, I am going to skip this, because I need to be working on logs and matricies :P

40. anonymous

It's fine. I'll type out what I did. First group: $(8x ^{3}+12x ^{2})+(-10x - 15)$ Then factor: $4x ^{2}(2x+3) - 5(2x+3)$ since the factored part (2x+3) is the same, you pair the multiplying factors and get this: $(4x ^{2}-5)(2x+3)$ Which works, and is in it's simplest form (since 4x^2-5 does not factor further as an integer).

41. anonymous

Good luck. :)

42. anonymous

tyvm!

43. anonymous

Welcome - thanks for sticking it out. It was going to drive me crazy! ;)

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