# (6-x)/(x^4+2x^3-13x^2-14x+24)+1/(x^3-2x^2-5x+6)+1/(x^2+x-2)-1/(x^2-4x+3)

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In order to sum the expression above, we'll try to find out the same denominator of all 4 ratios.

First of all, we'll try to find out the roots of the each denominator, in order to re-write them as products of linear factors.

We'll start with the simple denominator, meaning the second grade equations:

x^2-4x+3=0

x1={-(-4)+ sqrt[(-4)^2)-4*(1)(3)]}/2

**x1=6/2=3**

x2={-(-4)- sqrt[(-4)^2)-4*(1)(3)]}/2

**x2=2/2=1**

So, we could write the denominator using it's roots:

E(x)=(x-1)(x-3)

x^2+x-2=0

x1=[-1+sqrt(1+8)]/2

**x1=2/2=1**

x2=[-1-sqrt(1+8)]/2

**x2=-4/2=-2**

So, we could write the denominator using it's roots:

E(x)=(x-1)(x+2)

After that, we'll try the found roots in the first denominator, and if by substituting the root in denominator, the last one is equal to zero, then the value substituted is a root for this denominator, too.

Let's try with the value x1=1

The denominator is :E(x)=x^4+2x^3-13x^2-14x+24

After substitution the expression will become:

1^4+2*(1^3)-13*(1^2)-14*1+24=0

0=0, that means that x=1 is a root for the expression.

We'll do the same thing with the value x2=3

3^4+2*(3^3)-13*(3^2)-14*3+24=0

0=0

We'll write the expression using it's roots:

E(x)=x^4+2x^3-13x^2-14x+24 will become

E(x)=(x-1)(x-3)(x+2)(x+4)

We'll repeat the substitution in the next denominator:

E(x)=x^3-2x^2-5x+6

E(x)=1^3-2*(1^2)-5*1+6=0

E(x)=3^3-2*(3^2)-5*3+6=0

E(x)=(x-1)(x-3)(x+2)

Finally, we can write again the sum, in this way:

(6-x)/(x-1)(x-3)(x+2)(x+4)+1/(x-1)(x-3)(x+2)+1/(x-1)(x+2)-1/(x-1)(x-3)

It is obvious that we have to have the same denominator, in order to find out the resulat of the sum. The same denominator is (x-1)(x-3)(x+2)(x+4). We'll amplify each fraction with the missing paranthesis, so, the second fraction is amplified with (x+4), the third fraction with (x-3)(x+4) and the last fraction with (x+2)(x+4).

The result will appear in this way:

[(6-x)+(x+4)+(x-3)(x+4) -(x+2)(x+4)]/ (x-1)(x-3)(x+2)(x+4)=

[(6-x)+(x+4)(1+x-3-x-2)]/(x-1)(x-3)(x+2)(x+4)=

(6-x-4x-16)/(x-1)(x-3)(x+2)(x+4)=

(-10-5x)/(x-1)(x-3)(x+2)(x+4)=

-5(x+2)/(x-1)(x-3)(x+2)(x+4)=

**-5/(x-1)(x-3)(x+4)**

### (6-x)/(x^4+2x^3-13x^2-14x+24)+1/(x^3-2x^2-5x+6)+1/(x^2+x-2)-1/(x^2-4x+3).

Consider the denominator of first two terms:

### (x^4+2x^3-13x^2-14x+24)=(x^3-2x^2-5x+6)*(x+4).

Therefore their sum: (6-x)+(x+4)/(x^3-2x^2-5x+6)(x+4).

=10/(x^3-2x^2-5x+6)(x+4).

=10/(x-3)(x-1)(x+2)(x+4)...........................(1), as x^3-2x^2-5x+6 has its at x=1, 3 and -2.

Third term and 4th term:

### 1/(x^2+x-2)-1/(x^2-4x+3)

=[(x^2-4x+3)-(x^2+x-2)]/(x^2+x-2)(x^2-4x+3)

=-5(x-1)/{x+2)(x-1)(x-3)(x-1)}

=-5/{x+2)(x-3)(x-1)}

=-5(x+4)/(x-3)(x-1)(x+2)(x+4).............................(2)

Therefore (1)+(2) gives:

[10-5(x+4)]/(x-3)(x-1)(x+2)(x+4)

= (-5x-10)/(x-3)(x-1)(x+2)(x+4)

= -5(x+2)/(x-3)(x-1)(x+2)(x+4)

=-5/(x-3)(x-1)(x+4)