Respuesta :

suku54

Answer:

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =

         A2 - AB + BA - B2 =

         A2 - AB + AB - B2 =

         A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 16 is the square of 4

Check :  x4  is the square of  x2 

Factorization is :       (x2 + 4)  •  (x2 - 4) 

MrRoyal

Equivalent expressions are expressions with equal values

The equivalent expression of [tex]\mathbf{x^6 - 16x^2}[/tex] is [tex]\mathbf{x^2(x + 2)(x - 2)(x^2 + 4)}[/tex]

The expression is given as:

[tex]\mathbf{x^6 - 16x^2}[/tex]

Factor out x^2

[tex]\mathbf{x^6 - 16x^2 = x^2(x^4 - 16)}[/tex]

Express x^4 as (x^2)^2

[tex]\mathbf{x^6 - 16x^2 = x^2((x^2)^2 - 16)}[/tex]

Express 16 as 4^2

[tex]\mathbf{x^6 - 16x^2 = x^2((x^2)^2 - 4^2)}[/tex]

Apply difference of two squares

[tex]\mathbf{x^6 - 16x^2 = x^2((x^2 - 4)(x^2 + 4))}[/tex]

Express 4 as 2^2

[tex]\mathbf{x^6 - 16x^2 = x^2((x^2 - 2^2)(x^2 + 4))}[/tex]

Apply difference of two squares

[tex]\mathbf{x^6 - 16x^2 = x^2(x + 2)(x - 2)(x^2 + 4)}[/tex]

Hence, the equivalent expression of [tex]\mathbf{x^6 - 16x^2}[/tex] is [tex]\mathbf{x^2(x + 2)(x - 2)(x^2 + 4)}[/tex]

Read more about equivalent expressions at:

https://brainly.com/question/15715866