Por favor ayuda para resolver esta tarea es para hoy colocar en el paréntesis la letra que corresponda al producto de la columna de los factores?

Se colocan los resultados de las multiplicaciones de los binomios.

A) (x + 5)(x – 5) = (x)(x) + (x) – 5) + (5)(x) + (5)(– 5)

x² – 5x + 5x – 25

x² – 25

B) (x + 5)(x + 5) = (x)(x) + (x)(5) + (5)(x) + (5)(5)

x² + 5x + 5x + 25

x² + 10x + 25

C) (x + 5)(x – 3) = (x)(x) + (x)(– 3) + (5)(x) + (5)(– 3)

x² – 3x + 5x – 15

x² + 2x – 15

D) (x + 3)(x – 5) = (x)(x) + (x)(– 5) + (3)(x) + (3)(– 5)

x² – 5x + 3x – 15

x² – 2x – 15

E) (x – 5)(x – 5) = (x)(x) + (x)(– 5) + (– 5)(x) + (– 5)(– 5)

x² – 5x – 5x + 25

x² – 10x + 25

F) (– x – 5)(x – 5) = (–x)(x) + (– x)(– 5) + (– 5)(x) + (– 5)( – 5)

– x² + 5x – 5x + 25

– x² + 25

G) (– x + 3)( – x – 3) = (– x)(– x) + (– x)(– 3) + (3)(– x) + (3)(– 3)

x² + 3x – 3x – 9

x² – 9

H) (x + 5)(x + 3) = (x)(x) + (x)(3) + (5)(x) + (5)(3)

x² + 3x + 5x + 15

x² + 8x + 15

I) (x – 3)(x – 5) = (x)(x) + (x)(– 5) + (– 3)(x) + (– 3)(– 5)

x² – 5x – 3x + 15

x² – 8x + 15

J) (– x + 3)(x + 3) = (– x)(x) + (– x)(3) + (3)(x) + (3)(3)

– x² – 3x + 3x + 9

– x² + 9

K) (x + 2)(x – 2) = (x)(x) + (x)(– 2) + (2)(x) + (2)(– 2)

x² – 2x + 2x – 4

x² – 4

L) (x + 5)(x – 2) = (x)(x) + (x)(– 2) + (5)(x) + (5)(– 2)

x² – 2x + 5x – 10

x² + 3x – 10

M) (x – 2)(x – 2) = (x)(x) + (x)(– 2) + (– 2)(x) + (– 2)( – 2)

x² – 2x – 2x + 4

x² – 4x + 4

N) (– x + 2)( – x + 3) = (– x)(x) + (– x)(3) + (2)( – x) + (2)(3)

– x² – 3x – 2x + 6

– x² – 5x + 6

O) (x + 2)(x + 5) = (x)(x) + (x)(5) + (2)(x) + (2)(5)

x² + 5x + 2x + 10

x² + 7x + 10.