Mathematics, 25.07.2019 12:10 tashakelly42

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

Answers: 1

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Answers

Answer from: tayybabyy8

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

Answer from: Quest

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

$\frac{ab}{bc} =\frac{ad}{ab}$

now, we can cross multiply

and we get

$ab\timesab=ad\times bc$

$ab^2=ad\times bc$

so, this could be step to prove it

Seth is using the figure shown below to prove pythagorean theorem using triangle similarity: in the

Answer from: Quest

the answer is 8.64

step-by-step explanation:

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Consider the polynomial p(x)=32x^5y-2xy^5
part a: what is the complete factorization of p(x)=...

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