Question

Simplify completely:

6x^2+5x+1
-----------
27x^3+1

Answer 1

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

cuberoot 27= 3
cuberoot x^3=x
cuberoot 1=1

Answer 2

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.

Answer 3

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

Answer 4

D = b^2 - 4ac

Answer 5

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

Answer 6

yeah thats the denominator.

Answer 7

Or:
\[(3x+1)(9x^2-3x+1)\]

Answer 8

yep

Answer 9

So the numerator just stays the same then?

Answer 10

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)

Answer 11

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

Answer 12

Well, one sec...

Answer 13

it will. D = 1.

25- (6*4)

Answer 14

\[(6x-1)(x-1) = 6x^2-6x-1x+1 = 6x^2-7x+1\]

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

\[6x^2+5x+1\]

That means you have to use + signs to factor from what I have been told.

Answer 19

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

Answer 20

What is D then? The entire equation?

Answer 21

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

Answer 22

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

?

Answer 23

exactly..

Answer 24

\[-5 \pm \sqrt{25-24} \div 12\]

Answer 25

\[-5 \pm 1 \div 12\]

Answer 26

-6
----
12

or

-4
----
12

Answer 27

-2
---
4

-1
---
12

Answer 28

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

Answer 29

-1/3 and -1/2

Answer 30

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

Answer 31

Yes, I typed them wrong, I see that

Answer 32

so

-1/2 or -1/3

Answer 33

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

Answer 34

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

Answer 35

sorry it doesnt equal 0

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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