anonymous
  • anonymous
Simplify completely: 6x^2+5x+1 ----------- 27x^3+1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I know the 27x^3+1 is a perfect cube: cuberoot 27= 3 cuberoot x^3=x cuberoot 1=1
anonymous
  • 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
  • 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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anonymous
  • anonymous
D = b^2 - 4ac
anonymous
  • anonymous
so (3x+1)(9x^2-3x+1^2)?
anonymous
  • anonymous
yeah thats the denominator.
anonymous
  • anonymous
Or: \[(3x+1)(9x^2-3x+1)\]
anonymous
  • anonymous
yep
anonymous
  • anonymous
So the numerator just stays the same then?
anonymous
  • 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
  • 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
  • anonymous
Well, one sec...
anonymous
  • anonymous
it will. D = 1. 25- (6*4)
anonymous
  • anonymous
\[(6x-1)(x-1) = 6x^2-6x-1x+1 = 6x^2-7x+1\]
anonymous
  • anonymous
That doesn't match the equation. Where did you get a 4?
anonymous
  • anonymous
the formula i gave you is b^2 - 4ac. hence the 4..
anonymous
  • anonymous
But you can't use negatives when both signs in the equation are positive.
anonymous
  • anonymous
\[6x^2+5x+1\] That means you have to use + signs to factor from what I have been told.
anonymous
  • 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
  • anonymous
What is D then? The entire equation?
anonymous
  • anonymous
no..i told you D is the discriminant of the equation which is equal to B^2 - 4AC
anonymous
  • anonymous
So \[-5 \pm \sqrt{5^2-4(6)(1)} \div 2(6)\] ?
anonymous
  • anonymous
exactly..
anonymous
  • anonymous
\[-5 \pm \sqrt{25-24} \div 12\]
anonymous
  • anonymous
\[-5 \pm 1 \div 12\]
anonymous
  • anonymous
-6 ---- 12 or -4 ---- 12
anonymous
  • anonymous
-2 --- 4 -1 --- 12
anonymous
  • anonymous
But they are fractions. How am I supposed to plug those into the equation?
anonymous
  • anonymous
-1/3 and -1/2
anonymous
  • anonymous
the equation can be written as (x+1/3)(x+1/2) = 0.
anonymous
  • anonymous
Yes, I typed them wrong, I see that
anonymous
  • anonymous
so -1/2 or -1/3
anonymous
  • anonymous
But how does it help if it equals 0? That would solve for x for the top equation, but I am simplifying.
anonymous
  • anonymous
I thought the quadratic formula was for determining the values of x?
anonymous
  • anonymous
sorry it doesnt equal 0
anonymous
  • anonymous
its just that the numerator can be re written as (x+1/3)(x+1/2)
anonymous
  • anonymous
And also, (x+1/3_)x+1/2) \[\neq\] to \[6x^2+5x+1\]
anonymous
  • anonymous
Sorry, *(x+1/3)(x+1/2) \[\neq 6x^2+5x+1\]
anonymous
  • anonymous
Thats why I was saying it can't be factored.
anonymous
  • anonymous
it is equal to that divided by 6. you can always multiply and divide 6 and retain the eqn.
anonymous
  • anonymous
But it is already in a fraction. I can't have a fraction over a fraction and call it simplified.
anonymous
  • anonymous
Okay, figured something out, but didn't need quadratic formula for it.
anonymous
  • anonymous
\[\frac{(1+2 x)}{(1+6 x)^2} \]
anonymous
  • 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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