Factor the following expression ?

Question

X3+(2a+1)x2+(a2+2a-1)x+(a2-1)=

Answer 1

x³ + (2a + 1)x² + (a² + 2a - 1)x + (a² - 1) =

expand it:

x³ + 2ax² + x² + a²x + 2ax - x + a² - 1 =

group the terms as:

(x³ + 2ax² + a²x) + (x² + 2ax + a²) - (x + 1) =

factor out x from the first group :

x(x² + 2ax + a²) + (x² + 2ax + a²) - (x + 1) =

group the terms furtherly as:

[x(x² + 2ax + a²) + (x² + 2ax + a²)] - (x + 1) =

factor out (x² + 2ax + a²) from the first group as:

[(x² + 2ax + a²)(x + 1)]  - (x + 1) =

factor out the common term (x + 1):

(x + 1) [(x² + 2ax + a²) - 1] =

note that the trinomial within parentheses is a square, thus factorable as:

(x + 1) [(x + a)² - 1] =

finally, you can furtherly factor [(x + a)² - 1] as a difference between

two squares [(x + a)² - 1²], yielding:

(x + 1) [(x + a) + 1][(x + a) - 1] =

(x + 1)(x + a + 1)(x + a - 1)

thus, in conclusion:

x³ + (2a + 1)x² + (a² + 2a - 1)x + (a² - 1) = (x + 1)(x + a + 1)(x + a - 1)

I hope it helps...

Bye!

Answer 2

La factorización es la descomposición de un objeto en el producto de otros objetos más pequeños (factores), que, al multiplicarlos todos, resulta el objeto original.

Para factorizar esa expresión tienes que hallar sus 3 soluciones:

X³+(2a+1)x²+(a²+2a-1)x+(a²-1) = (x - S1) (x - S2) (x - S3)

operando:

(x - S1) (x - S2) (x - S3) = x³ - (S1 + S2 +S3)x² + (S1S2 + (S1 + S2)S3)x - S1S2S3 = x³ - (S1 + S2 +S3)x² + (S1S2 + S1S3 + S2S3)x - S1S2S3

Ahora identificamos:

2a+1 = -(S1 + S2 + S3) = - S1 - S2 - S3

a²+2a-1 = S1S2 + S1S3 + S2S3

a²-1 = -S1S2S3 = (a+1)(a-1)

y falta resolver este sistema de 3 ecuaciones con 3 incógnitas.

Supongamos que S3 = -1

2a+1 = - S1 - S2 + 1

a²+2a-1 = S1S2 - S1 - S2

a²-1 = S1S2 = (a+1)(a-1)

De la tercera: a²-1 = S1S2 = (a+1)(a-1)

S1 = -(a+1)

S2 = -(a-1)

En la primera: - S1 - S2 + 1 = -[-(a+1)] - [-(a+1)] + 1 = = a +1 + a -1 + 1 = 2a +1 Se cumple (y es la que dice que van los signos menos en el anterior apartado.

Falta verificar la segunda:

S1S2 - S1 - S2 = [-(a+1)] [-(a-1)] - [-(a+1)] - [-(a-1)]

= (a+1)(a-1) +(a+1) + (a-1) = a² -1 +2a = a² +2a -1 y tambien se verifica.

Solución:

X³+(2a+1)x²+(a²+2a-1)x+(a²-1) = (x - S1) (x - S2) (x - S3) =

= (x - [-(a+1))] (x - [-(a-1))] [x - (-1)] =

= (x + (a+1)) (x + (a-1)) (x + 1)

Salu2