Answer:
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
& 
of this 
