\(\frac{{4x}^{2}+3}{{x}^{3}+{x}^{2}+x}\)
That is the original problem. I've gotten this far with it.
\(\frac{{4x}^{2}+3}{x({x}^{2}+x+1)}\)=\(\frac{A}{x}+\frac{Bx+C}{{x}^{2}+x+1}\)
and
\((A)({x}^{2}+x+1)+(Bx+C)(x)=4{x}^{2}+3\)
And since its true for all x values, I can plug in x=0 and then I know that
A=3 .
But I don't know how to find B and C.
Does anyone know how to do this? I watched 3 Khan academy videos on it and I looked at the examples in my book and I'm pretty stuck. :(
+129839 A ( x^2 + x + 1) + (Bx + C) x = 4x^2 + 3
Ax^2 + Ax + A + Bx^2 + Cx = 4x^2 + 3
(A + B)x^2 + ( A + C)x + A = 4x^2 + 0x + 3 equate coefficients
A + B = 4
A + C = 0
A = 3 → C = - 3 → B = 1
Proof
3 / x + [1x - 3] / [ x ^2 + x + 1] =
[3 (x^2 + x + 1) + x (1x - 3) ] / [ x ( x^2 + x + 1 ] =
[ 3x^2 + 3x + 3 + 1x^2 - 3x ] / [ x^3 + x^2 + x] =
[4x^2 + 3] / [x^3 + x^2 + x]
