1.)The polynomial 729 x^{3} - 1 y^{3} can be factored into the product of two polynomials, A * B where the degree of A is greater than the degree of B. Find A and B.
2.)The polynomial 16 a^{6} + 8 a^{3}b + 1 b^2 - 49 c^{6} 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.
strange wording for a problem that simply could be stated as ...
factor 729x^3 - y^3
then
729x^3 - y^3 = (9x-y)(81x^2 + 9xy + y^2)
so I guess
B is 9x-y and
A is 81x^2 + 9xy + y^2
For the second, I will simply factor it, ignoring the strange conditions.
16a^6 + 8a^2b + b^2 - 49c^6
= (4a^2+b)^2 - 49c^6 , a difference of squares
= (4a^2 + b + 7c^3)(4a^2 + b - 7c^3)
wow thank you so much! i spent hours on that problem. thanks again for the help.
I am just curious who is giving you such unusually phrased questions.
Are these from your textbook ?
It's from a website called WebWork where we do the homework on.
To find the factors of the given polynomials, we can use the difference of cubes formula for the first question and factor by grouping for the second question.
Question 1:
The given polynomial is 729x^3 - 1y^3.
We can use the difference of cubes formula, which states that a^3 - b^3 can be factored as (a - b)(a^2 + ab + b^2).
By applying this formula, we can rewrite the polynomial as:
(9x - y)(81x^2 + 9xy + y^2).
So, the factors of the given polynomial are A = (9x - y) and B = (81x^2 + 9xy + y^2), where the degree of A is greater than the degree of B.
Question 2:
The given polynomial is 16a^6 + 8a^3b + b^2 - 49c^6.
To factor this polynomial, we can use the technique of factor by grouping.
We group the terms and factor them separately:
(16a^6 + 8a^3b) + (b^2 - 49c^6).
For the first group, we can factor out common terms:
8a^3(2a^3 + b).
For the second group, we can recognize it as a difference of squares:
(b - 7c^3)(b + 7c^3).
So, A = 8a^3(2a^3 + b) and B = (b - 7c^3)(b + 7c^3), where the coefficient of c in A is less than the coefficient of c in B.