Question

resuelve las multiplicaciones de polinomios

necesito ayuda pliss doy coronita es para hoy!! ​

Answer 1

Explicación:

a) (2x²)•(3x) =

[tex] = (2)(3)x^{2 + 1} \\ = 6 {x}^{3} [/tex]

b) (2x+1)•(3x+2)=

[tex] = (2x)(3x) + (2x)(2) + (1)(3x) + (1)(2) \\ = 6 {x}^{1 + 1} + 4x + 3x + 2 \\ = 6 {x}^{2} + 7x + 2[/tex]

c) (3x + 3)•(x² + 2x + 1) =

[tex] = (3x)( {x}^{2} ) + (3x)(2x) + (3x)(1) + (3)( {x}^{2} ) + (3)(2x) + (3)(1) \\ = 3 {x}^{1 + 2} + 6 {x}^{1 + 1} + 3x + 3 {x}^{2} + 6x + 3 \\ = 3 {x}^{3} + 6 {x}^{2} + 3 {x}^{2} + 9x + 3 \\ = 3 {x}^{3} + 9 {x}^{2} + 9x + 3[/tex]

d) 3•(2x³ - 3x² + 4x - 2) =

[tex] = (3)(2 {x}^{3} ) - (3)(3 {x}^{2} ) + (3)(4x) - (3)(2) \\ = 6 {x}^{3} - 9 {x}^{2} + 12x - 6[/tex]

e) 3x²• (2x³- 3x² + 4x - 2) =

[tex] = (3 {x}^{2} )(2 {x}^{3} ) - (3 {x}^{2} )(3 {x}^{2} ) + (3 {x}^{2} )(4x) - (3 {x}^{2} )(2) \\ = 6 {x}^{2 + 3} - 9 {x}^{2 + 2} + 12 {x}^{2 + 1} - 6 {x}^{2} \\ = 6 {x}^{5} - 9 {x}^{4} + 12 {x}^{3} - 6 {x}^{2} [/tex]