Consider the polynomial p (x) = 32x^5y-2xy^5

part a: What is the complete factorization of p (x) = 32x^5y-2x^5 over the integers?

part b: what methods are used to factor p (x) = 32x^5y^5?

Select 1 answer for a and one for b.

a - 2xy (2x-y) ^2 (2x+y) ^2

a-2xy (2x-y) (2x+y) (4x^2+y^2

a-2xy (4x^2-y^2) (x^4-4x^2y^2+y^4

b-repeated differences of squares

b - difference of cubes

b-greatest common factor

b-grouping

a - p (x) = 2xy (2x - y) (2x + y) (4x² + y²) ⇒ 2nd answer

b - repeated differences of squares ⇒ 1st answer

Step-by-step explanation:

* Lets explain how to solve the problem

∵ p (x) = 32x^5y - 2xy^5

- The coefficients of the two terms are 32 and 2

∵ 2 is a common factor in 32 and 2

∵ 32 : 2 = 16 and 2 : 2 = 1

∴ p (x) = 2 (16x^5y - xy^5)

* Now lets find the common factors of the variables x and y

∵ The common factor is x^5 and x is x

∵ The common factor in y and y^5 is y

∴ the common factors in 16x^5y - xy^5 are xy

∵ 16x^5y : xy = 16x^4

∵ xy^5 : xy = y^4

∴ p (x) = 2xy (16x^4 - y^4)

- Remember that a² - b² is called difference of two squares we

factorize it by distributed into two polynomials have same terms

with different middle sign (a + b) (a - b)

∵ 16x^4 - y^4 is a different of two squares because √ (16x^4) = 4x²

and √9y^4) = y²

∴ The factorization of 16x^4 - y^4 is (4x² + y²) (4x² - y²)

∴ p (x) = 2xy[ (4x² + y²) (4x² - y²) ]

- The bracket 4x² - y² is also different of two squares because

√ (4x²) = 2x and √ (y²) = y

∴ The factorization of 4x² - y² is (2x - y) (2x + y)

∴ p (x) = 2xy (2x - y) (2x + y) (4x² + y²)

a - p (x) = 2xy (2x - y) (2x + y) (4x² + y²)

b - The methods used to factor p (x) are:

greatest common factor and repeated differences of squares

- But you ask to chose one answer of b so chose repeated

differences of squares