Respuesta :

ItzMikeSS

Answer: [tex]4a^4+4b^4-8a^3b+8ab^3-13a^2b^2[/tex]

Step-by-step explanation:

[tex]4(a^2-b^2)^2-8ab(a^2-b^2)-5a^2b^2[/tex]

First, solve the square binomial.

[tex]4((a^2)^2-2(a^2)(b^2)+(b^2)^2)-8ab(a^2-b^2)-5a^2b^2[/tex]

[tex]4(a^4-2a^2b^2+b^4)-8ab(a^2-b^2)-5a^2b^2[/tex]

Now, distribute the 4 and 8ab respectively.

[tex]4a^4-8a^2b^2+4b^4-8a^3b+8ab^3-5a^2b^2[/tex]

Combine like terms. (I'll group them so that you can see them more clearly)

[tex]4a^4+4b^4-8a^3b+8ab^3+(-8a^2b^2-5a^2b^2)[/tex]

[tex]4a^4+4b^4-8a^3b+8ab^3+(-13a^2b^2)[/tex]

[tex]4a^4+4b^4-8a^3b+8ab^3-13a^2b^2[/tex]

mhanifa

Answer:

(2a²-2b²+a²b²)(2a²-2b²-5a²b²)

Step-by-step explanation:

4(a²-b²)²-8ab(a²-b²)-5a²b²

Let's replace 2(a²-b²)= m and a²b²=n for simplicity

then we have

  • m²-4mn-5n²
  • = m²-4mn+4n²-4n²-5n²
  • = (m-2n)²- 9n²
  • = (m-2n)²- (3n)²
  • = (m-2n+3n)(m-2n-3n)
  • = (m+n)(m-5n)

Now we can replace m and n with initial values:

  • (m+n)(m-5n)=
  • = (2(a²-b²)+a²b²)(2(a²-b²)- 5a²b²)
  • = (2a²-2b²+a²b²)(2a²-2b²-5a²b²)

Otras preguntas