# Prove that there are three real numbers a and b and c so that: $(\forall x\ne -1/2)\frac{6x^2+7x-3}{2x+1}=ax+b+\frac{c}{2x+1}$

1) Prove that there are three real numbers a and b and c so that: $(\forall x\ne -1/2)\frac{6x^2+7x-3}{2x+1}=ax+b+\frac{c}{2x+1}$

[[Addition: this question 1) is related to this question 2):

Conclude from 1) the value of this integral: $\int_0^1\frac{6x^2+7x-3}{2x+1}dx$]]

I never seen question like this in my life, how can we ever prove this? (I'm talking about 1) )

Thank you!!

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Reduce the RHS to the same denominator and identify the coefficients of the numerator. – user63181 Dec 16 '13 at 11:49
What? What is RHS? I didn't understand what you mean by that. – Wolf Dec 16 '13 at 11:51
$6x^2+7x-3=(2x+1)(3x+2)-5$ – mathlove Dec 16 '13 at 11:52

$$ax+b+\frac{c}{2x+1}=\frac{(ax+b)(2x+1)+c}{2x+1}=\frac{2ax^2+(a+2b)x+b+c}{2x+1}=\frac{6x^2+7x-3}{2x+1}$$ so what's the value of $a,b$ and $c$?
The polynomial $2ax^2+(a+2b)x+b+c$ is equal to $6x^2+7x-3$ so the coefficient of $x^2$: $2a=6$ and the coefficient of $x$: $a+2b=7$ and the constant $b+c=-3$. Solve these equations. – user63181 Dec 16 '13 at 12:04