Ask Question
6 September, 15:11

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

+3
Answers (1)
  1. 6 September, 15:45
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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: ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers