MHB C.c.'s question at Yahoo Answers regarding factoring a polynomial

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The polynomial 81a^{14} - 18a^{7}b + b^{2} - 16c^{-4} can be factored using the difference of squares method. It is first expressed as the difference of two squares: (9a^7 - b)² - (4c^{-2})². This leads to the factorization into two polynomials: (9a^7 - b + 4c^{-2})(9a^7 - b - 4c^{-2}). The polynomial A is chosen to have a smaller coefficient for c, resulting in A = 9a^7 - b - 4c^{-2} and B = 9a^7 - b + 4c^{-2}. The solution provides a clear step-by-step approach to factoring the polynomial.
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Here is the question:

Help with math question please!?

The polynomial 81a^{14} - 18a^{7}b + 1b^{2} - 16c^{-4} can be factored into the product of two polynomials, A & B, where the coefficient of c in A is less than the coefficient of c in B. Find A and B.

If you could, try to show me how to solve this step by step.
{exponents}

Here is a link to the question:

Help with math question please!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello c.c.,

We are given to factor:

$$81a^{14}-18a^{7}b+b^{2}-16c^{-4}$$

We may factor the first 3 terms as the square of a binomial, while the 4th term is a square as well:

$$\left(9a^7-b \right)^2-\left(4c^{-2} \right)$$

Using the difference of squares formula $$x^2-y^2=(x+y)(x-y)$$, we may write the expression as:

$$\left(9a^7-b+4c^{-2} \right)\left(9a^7-b-4c^{-2} \right)$$

Now, choosing $A$ such that the coefficient on the term containing $c$ is the smaller one, we have:

$$A=9a^7-b-4c^{-2}$$

$$B=9a^7-b+4c^{-2}$$

To c.c., and any other guests viewing this topic, I invite and encourage you to register and post other factoring questions in our http://www.mathhelpboards.com/f2/ forum.

Best Regards,

Mark.
 
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