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

MarkFL

Staff member
Here is the question:

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:

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

MarkFL

Staff member
Hello c.c.,

We are given to factor:

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

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

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

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

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

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

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

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

To c.c., and any other guests viewing this topic, I invite and encourage you to register and post other factoring questions in our Pre-Algebra and Algebra forum.

Best Regards,

Mark.