1) x^8+x^4+1 2)a^3-3a+2 3) x^4+x^2y^2+y^4 4) x^3+3x^2-4 - вопрос №1841233234224433245814519 от kashavcevaeliz 15.04.2022 09:34

Question

Алгебра: 1) x^8+x^4+1 2)a^3-3a+2 3) x^4+x^2y^2+y^4 4) x^3+3x^2-4

kashavcevaeliz · Answer

1) x⁸+x⁴+1 = (x⁴)² + 2x⁴ +1² - x⁴ = (x⁴+1)² - (x²)² = (x⁴+1-x²)(x⁴+1+x²) =
= (x⁴-x²+1)(x⁴+x²+1) = (x⁴-x²+1) ( (x²)² +2x²+1² - x² ) =
= (x⁴ -x²+1) ( (x²+1)² - x² ) = (x⁴-x²+1)(x²+1-x )(x²+1+x)=
= (x⁴ -x²+1)(x²-x+1)(x²+x+1)

2) a³-3a+2 = a³ - 1³ -3a+3 = (a-1)(a²+a+1) -3(a-1) =
= (a-1)(a²+a+1-3) = (a-1)(a²+a-2)= (a-1)(a²+2a-a-2) =
= (a-1)(a(a+2) -1(a+2)) = (a-1)(a-1)(a+2) =
= (a-1)²(a+2)

3)x⁴+x²y²+y⁴= (x²)² + 2x²y² +(y²)² - (xy)²=
= (x²+y²)² - (xy)² = (x² +y²- xy)(x²+y²+xy)

4)x³+3x²-4 = x³ +2x² + x² - 2² = x²(x+2) +(x+2)(x-2) =
=(x+2)(x² +x-2) = (x+2)(x² -1 +x-1) = (x+2)( (x-1)(x+1) +1(x-1)) =
= (x+2)(x-1)(x+1+1) = (x+2)²(x-1)