Find Independent of x in (6x^2-3/x)^6 Using Pascal's Triangle

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In summary, the conversation discusses finding the independent coefficient or constant term in the expansion of the expression (6x^2-\frac{3}{x})^6. The user is unsure about the term "independent of x" and is seeking clarification. They provide their working and arrive at the answer of r = 4. The conversation ends with a thank you and mention of the end of exams.
  • #1
Mathysics
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how do i find the independent of x in (6x^(2)-3/x)^6

I know that the formula nCr a^(n-r) b^r

i have got an answer of n = 6 and r = 4 i just want to make sure that I am correct and i would like someone to give me a hand~ :)
 
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  • #2
I don't think people (including myself) understand what you mean by finding "the independent of x".
 
  • #3
Do you mean possibly the "constant term" (which is "independent of x") when you expand the polynomial? If so, then yes, r= 4 is correct. However, in future, it would be good to show how you got your answer so (just in case you are wrong!) we could indicate where you went wrong.
 
  • #4
I believe the sought expression is "coefficient". As in: Find the coefficient of x in the expansion of the expression [itex](6x^2-\frac{3}{x})^6[/itex].

Obviously n = 6 here. In this case one would need to find what value of r results in [itex]a^r b^{(n-r)}[/itex] being a multiple of x (here [itex]a=6x^2,b=-\frac{3}{x}[/itex].

--Elucidus
 
  • #5
loveequation said:
I don't think people (including myself) understand what you mean by finding "the independent of x".
independent of x simply means that x^0 and yes I am trying to find the co-efficient there

heres are my working out.
independent of x in (6x^(2)-3/x)^6

1. (6^(6-r) x^(12-2r) ((-3^(r))/x^(r))

2. x^(12-2r)/x^(r) = x^(0)

3. x^(12-3r)=x^(0)

4. 12-3r=0

5. r = 4
 
  • #6
Mathysics said:
independent of x simply means that x^0 and yes I am trying to find the co-efficient there

heres are my working out.
independent of x in (6x^(2)-3/x)^6

1. (6^(6-r) x^(12-2r) ((-3^(r))/x^(r))

2. x^(12-2r)/x^(r) = x^(0)

3. x^(12-3r)=x^(0)

4. 12-3r=0

5. r = 4

Ah, I see. This is a term I was unfamiliar with, but I understand now. Your work seems correct so far. If r = 4, what would the independent coefficient (constant term) be?

--Elucidus
 
  • #7
thx for the help :)

exams finally over
 

1. How do you use Pascal's Triangle to find the independent term of a polynomial?

The independent term of a polynomial is the term that does not contain any variables. To use Pascal's Triangle to find this term, we need to expand the given polynomial using the binomial theorem. The coefficients of the expanded polynomial will follow the pattern of Pascal's Triangle, with the independent term being the last term of the expansion.

2. Can you explain the steps to find the independent term in (6x^2-3/x)^6 using Pascal's Triangle?

First, we need to expand the given polynomial using the binomial theorem. The expansion will look like: (6x^2)^6 - 6(6x^2)^5(3/x) + 15(6x^2)^4(3/x)^2 - 20(6x^2)^3(3/x)^3 + 15(6x^2)^2(3/x)^4 - 6(6x^2)(3/x)^5 + (3/x)^6. The coefficients of each term follow the pattern of Pascal's Triangle. The independent term is the last term, which is (3/x)^6.

3. How does Pascal's Triangle help in finding the independent term of a polynomial?

Pascal's Triangle is a useful tool in finding the coefficients of a polynomial expansion. The coefficients follow a specific pattern, making it easier to find the independent term of a polynomial. By using this pattern, we can quickly determine the independent term without having to expand the entire polynomial.

4. Is it possible to find the independent term of a polynomial without using Pascal's Triangle?

Yes, it is possible to find the independent term of a polynomial by expanding the polynomial using the binomial theorem. However, using Pascal's Triangle can make the process faster and more efficient, especially for higher degree polynomials.

5. Can Pascal's Triangle be used for any polynomial, or only for specific types?

Pascal's Triangle can be used for any polynomial that can be expanded using the binomial theorem. This includes polynomials with both positive and negative exponents. However, for polynomials with more than two terms, using Pascal's Triangle may not be as efficient as using other methods such as the FOIL method or the distributive property.

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