# Use elementary fractions to write the fraction 1/(x^3+x^2+x+1)

justaguide | College Teacher | (Level 2) Distinguished Educator

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To rewrite 1/(x^3+x^2+x+1) as partial fractions, we first need to factorize the denominator

1/(x^3+x^2+x+1)

=> 1 / x^2(x + 1) + 1(x +1)

=> 1 / (x^2 + 1)(x + 1)

Let this be equal to (Ax +B) / (x^2 +1) + C / (x +1)

(Ax + B) / (x^2 +1) + C / (x +1) = 1 / (x^2 + 1)(x + 1)

=> (Ax + B) (x +1) + C(x^2 +1) = 1

=> Ax^2 + Bx +Ax + B + Cx^2 + C = 1

=> Ax^2 + Cx^2 = 0 , Ax + Bx = 0 , B + C = 1

=> A + C = 0 , A + B = 0 and B + C = 1

=> C = -A , B = -A and B + C = 1

=> -A + -A = 1

=> A = -1/2

C = 1/2 and B = 1/2

So (Ax +B) / (x^2 +1) + C / (x +1)

=> (-1/2)x +(1/2) / (x^2 +1) + (1/2) / (x +1)

=> (1 - x)/ 2(x^2 +1) + 1/2(x +1)

Therefore the required result is (1-x)/[2(x^2 +1)] +1/2*(x +1)

neela | High School Teacher | (Level 3) Valedictorian

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To use in elementary fractions to write the fraction 1/(x^3+x^2+x+1).

x^3+x^2+x+1 =  (x^2(x+1) +1.(x+1) = (x+1)(x^2+1).

Therefore  let 1/(x^2+x^2+x+1) = a/(x+1) + (bx+c)/(x^2+1)....(1).

We have to determine a, b and c.

We multiply both sides of (1) by (x+1)(x^2+1) and we get:

1 = a(x^2+1) + (bx+c)(x+1)

1 =  ax^2+a +bx^2+bx+cx+c.

1 = (a+b)x^2+(b+c)x+ (a+c). We treat this as an identity. So we can equate the coefficient of like terms on both sides:

x^2 terms: 0 = a+b...(1).

x terms = 0 = b+c......(2).

Constant terms: 1 = a+c, or a+c = 1.....(3)

(1) -(2): a - c = 0...(4).

(3)+(4) : 2a = 1. So a= 1/2.

(3-(4): 2c =  1. So c = 1/2.

Therfore we put a = 1/2 in (1) and get a+b = 0, or (1/2 +b = 0. S = b= -1/2.

Therefore a = 1/2, b=-1/2 and c = 1/2.

So

1/(x^3+x^2+x+1) = 1/{2(x+1)} + (-x+1)/{2(x^2+1)}

1/(x^3+x^2+x+1) = 1/(2(x+1)} +1/{2{x^2+1) - x/{2(x^2+1)}.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll decompose the given fraction into elementary ratios.

First, we'll factorize the denominator x^3+x^2+x+1. We'll group the first 2 terms and the last 2 terms.

(x^3+x^2) + (x+1) = x^2(x+1) + (x+1)

We'll factorize by (x+1):

(x^3+x^2+x+1) = (x+1)(x^2+1)

1/(x^3+x^2+x+1) = A/(x+1) + (Bx + C)/(x^2 + 1)

We'll multiply both sides by (x+1)(x^2+1):

1 = A(x^2 + 1) + (Bx + C)(x+1)

We'll remove the brackets:

1 = Ax^2 + A + Bx^2 + Bx + Cx + C

We'll combine like terms:

1 = x^2 (A + B) + x (B + C) + A + C

A + B = 0 => A = -B

B + C = 0 => C = -B => C = A

A + C = 1

2A = 1

A = C = 1/2

B = -1/2

The fraction 1/(x^3+x^2+x+1) could be written as:

1/(x+1)(x^2+1) = 1/2(x+1) - (x - 1)/2(x^2 + 1)