- anonymous

Simplify completely:
6x^2+5x+1
-----------
27x^3+1

- katieb

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- anonymous

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

- 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.

- 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

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## More answers

- anonymous

D = b^2 - 4ac

- anonymous

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

- anonymous

yeah thats the denominator.

- anonymous

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

- anonymous

yep

- anonymous

So the numerator just stays the same then?

- 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)

- 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

- anonymous

Well, one sec...

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

What is D then? The entire equation?

- anonymous

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

- anonymous

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

- anonymous

exactly..

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

-1/3 and -1/2

- anonymous

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

- anonymous

Yes, I typed them wrong, I see that

- anonymous

so
-1/2 or -1/3

- anonymous

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

- anonymous

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

- anonymous

sorry it doesnt equal 0

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

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