Replace ∗ with a monomial so that the result is an identity:

1. (2a + ∗)(2a - ∗) = 4a^2–b^2

2. (∗− 3x )(∗ +3x) = 16y^2–9x^2

3. 100m^4–4n^6 = (10m^2−∗)(∗ +10m^2)

4. m^4–225c^10 = (m^2−∗)(∗ +m^2).

Respuesta :

Answer:

Each of these is based on the principle of difference of two squares where:

[tex]a^2-b^2=(a-b)(a+b)[/tex]

Step-by-step explanation:

1.[tex](2a + *)(2a - *) = 4a^2-b^2[/tex]

[tex]4a^2-b^2=(2a)^2-b^2=(2a+b)(2a-b)[/tex]

2. [tex](*- 3x )(*+3x) = 16y^2-9x^2[/tex]

[tex]16y^2-9x^2=(4y)^2-(3x)^2=(4-3x)(4+3x)[/tex]

3. [tex]100m^4-4n^6 = (10m^2-*)(*+10m^2)[/tex]

[tex]100m^4-4n^6 = (10m^2-(2n^3)^2)= (10m^2-2n^3)(2n^3+10m^2)[/tex]

4. [tex]m^4-225c^{10} = (m^2-*)(* +m^2)[/tex]

[tex]m^4-225c^{10} =(m^2)^2-(15c^5)^2= (m^2-15c^5)(15c^5 +m^2)[/tex]