is an identity with respect to x

[Andrew08]

I can't find these 3 values to save my life

 

[MathLover1]

(x^2+1)/(x^3+x^2)=a/x+b/x^2+c/(x+1) where a,b,c are your numbers in box

first left side
(x^2+1)/(x^3+x^2)
=(x^2+1)/(x^2(x+1))

then right side
a/x+b/x^2+c/(x+1)=(ax^2 + ax + bx + b + cx^2)/(x^2 (x + 1))

then

(x^2+1)/(x^2(x+1))=(ax^2 + ax + bx + b + cx^2)/(x^2 (x + 1))...........if denominators same, for equation to be true, numerators must be same to

x^2+1=ax^2 + ax + bx + b + cx^2

x^2+0*x+1=(a+ c)x^2 + (a + b)x + b

=>a+c=1
=>(a + b)=0=>a=-b
=>b=1
would be true if: a = -1, b = 1, c = 2

 

[Andrew08]

(x^2+1)/(x^3+x^2)=a/x+b/x^2+c/(x+1) where a,b,c are your numbers in box

first left side
(x^2+1)/(x^3+x^2)
=(x^2+1)/(x^2(x+1))

then right side
a/x+b/x^2+c/(x+1)=(ax^2 + ax + bx + b + cx^2)/(x^2 (x + 1))

then

(x^2+1)/(x^2(x+1))=(ax^2 + ax + bx + b + cx^2)/(x^2 (x + 1))...........if denominators same, for equation to be true, numerators must be same to

x^2+1=ax^2 + ax + bx + b + cx^2

x^2+0*x+1=(a+ c)x^2 + (a + b)x + b

=>a+c=1
=>(a + b)=0=>a=-b
=>b=1
would be true if: a = -1, b = 1, c = 2

thank you so much, man!!

 

[MathLover1]

thank you so much, man!!

Mamm!

 

[Andrew08]

Mamm!

oh sorry

 

[MathLover1]

oh sorry

no problem