## anonymous 5 years ago Simplify completely: 6x^2+5x+1 ----------- 27x^3+1

1. anonymous

I know the 27x^3+1 is a perfect cube: cuberoot 27= 3 cuberoot x^3=x cuberoot 1=1

2. anonymous

but the top can't be factored. I think I can rewrite this as (3x+1)^3 But would that make the answer 6x^2+5x+1 ----------- ? (3x+1)^3 I can't help but think there is more to this.

3. anonymous

factorize the denominator..an equation in the form a^3 + b^3 = (a+b)(a^2-ab+b^2) also factorize the numerator. the numerator can be factorized by using the formula x = $-b \pm \sqrt{D} \div 2a$ where the quadratic is in the form ax^2 + bx + c

4. anonymous

D = b^2 - 4ac

5. anonymous

so (3x+1)(9x^2-3x+1^2)?

6. anonymous

yeah thats the denominator.

7. anonymous

Or: $(3x+1)(9x^2-3x+1)$

8. anonymous

yep

9. anonymous

So the numerator just stays the same then?

10. anonymous

no. factorize the numerator using the formula i gave you..the formula gives you the roots of the equation. if a is the 1st root and b is the 2nd root then the numerator can be re written as (x-a)(x-b)

11. anonymous

It won't work that way because there are 2 + signs, which means the signs must stay the same. Nothing but 1*1 will = +1

12. anonymous

Well, one sec...

13. anonymous

it will. D = 1. 25- (6*4)

14. anonymous

$(6x-1)(x-1) = 6x^2-6x-1x+1 = 6x^2-7x+1$

15. anonymous

That doesn't match the equation. Where did you get a 4?

16. anonymous

the formula i gave you is b^2 - 4ac. hence the 4..

17. anonymous

But you can't use negatives when both signs in the equation are positive.

18. anonymous

$6x^2+5x+1$ That means you have to use + signs to factor from what I have been told.

19. anonymous

nope..for any quadratic expression of the form $Ax ^{2} + Bx + C$ the formula for the roots are $-B \pm \sqrt{D} / 2A$

20. anonymous

What is D then? The entire equation?

21. anonymous

no..i told you D is the discriminant of the equation which is equal to B^2 - 4AC

22. anonymous

So $-5 \pm \sqrt{5^2-4(6)(1)} \div 2(6)$ ?

23. anonymous

exactly..

24. anonymous

$-5 \pm \sqrt{25-24} \div 12$

25. anonymous

$-5 \pm 1 \div 12$

26. anonymous

-6 ---- 12 or -4 ---- 12

27. anonymous

-2 --- 4 -1 --- 12

28. anonymous

But they are fractions. How am I supposed to plug those into the equation?

29. anonymous

-1/3 and -1/2

30. anonymous

the equation can be written as (x+1/3)(x+1/2) = 0.

31. anonymous

Yes, I typed them wrong, I see that

32. anonymous

so -1/2 or -1/3

33. anonymous

But how does it help if it equals 0? That would solve for x for the top equation, but I am simplifying.

34. anonymous

I thought the quadratic formula was for determining the values of x?

35. anonymous

sorry it doesnt equal 0

36. anonymous

its just that the numerator can be re written as (x+1/3)(x+1/2)

37. anonymous

And also, (x+1/3_)x+1/2) $\neq$ to $6x^2+5x+1$

38. anonymous

Sorry, *(x+1/3)(x+1/2) $\neq 6x^2+5x+1$

39. anonymous

Thats why I was saying it can't be factored.

40. anonymous

it is equal to that divided by 6. you can always multiply and divide 6 and retain the eqn.

41. anonymous

But it is already in a fraction. I can't have a fraction over a fraction and call it simplified.

42. anonymous

Okay, figured something out, but didn't need quadratic formula for it.

43. anonymous

$\frac{(1+2 x)}{(1+6 x)^2}$

44. anonymous

Sorry. The denominator should be: $\left(1-3 x+9 x^2\right)$ $\frac{6 x^2+5 x+1}{27 x^3+1}=\frac{1+2 x}{1-3 x+9 x^2}$