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I have the question

$$ \frac{ 3x + 3 }{ (x-1)(x^2 +x +1) } $$

and I am unsure about how to start as the quadratic on the denominator is irreducible. So anyone got any tips for starting this one?

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  • $\begingroup$ You can still use partial fractions. $\endgroup$ – James Sep 28 '14 at 21:47
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Write

$$\frac{3x+3}{(x-1)(x^2+x+1)} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+x+1)} $$

Multiply the equation by $(x-1)(x^2+x+1)$ and solve for A,B and C.

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Because the quadratic is irreducible, you need to use a non-constant term in the expansion.

So something of the form: $$\frac{3x+3}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$$

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  • $\begingroup$ It was that plus I knew $ BxC $ didn't look right $\endgroup$ – Paul Sep 28 '14 at 21:50
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Ok so start the partial fraction as (3x+3)/(x-1) (x^2+x+1) ={ A/(x-1) } + { (Bx+C)/(x^2+x+1)}

then solve for A, B and C by reducing the above equation 3x+3 = A (x^2+x+1) + B(x)(x-1) + C (x-1) 3x+3 = Ax^2 + Ax + A + Bx^2 - Bx + Cx-C equate all the x,x^2 and constant terms on the left and right side of the above eq 3x= x(A-B+C) 0= x^2 (A+B) 3= A -C solve these 3 linear equations in 3 variables and u can get the answer. for A B C then substitute in the original partial fraction. Good Luck

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