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

-4

step-by-step explanation:

(-4+8)=-(4-8)=-8+4=-4

25

step-by-step explanation: