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I'm stuck on a basic question regarding identities.

$A(x^2-1)+B(x-1)+C = (3x-1)(x+1)$

I've managed to substitute $x$ for $1$ to work out C is $4$. However, I'm unsure how to work out A and B respectively.

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up vote 0 down vote accepted

For $x=-1$ we have, $-2B+C=0\Rightarrow 2B=C=4\Rightarrow B=2$

for $x=0$ we have , $-A-B+C=-1 \Rightarrow A=1-B+C\Rightarrow A=3$

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can you explain further why you chose these values? – peter_gent Jun 11 '13 at 14:25
-1 because -1 is a factor of most of the expressions – Abhra Abir Kundu Jun 11 '13 at 14:25
can you please generalize why that is important or give another example ? :) – peter_gent Jun 11 '13 at 14:27
putting 0 is always helpful. Then look for the factor which is present most no. of times because it will help you to eliminate most of the factors. – Abhra Abir Kundu Jun 11 '13 at 14:31
if x is -1 then: -2B+4=-4 if we know that C is 4 ? – peter_gent Jun 11 '13 at 14:36

For A equate the coefficients of $x^2$ on each side. For $B$ set $x=-1$ which kills the term with $A$ in.

You can multiply through and equate coefficients to get three equations in three unknowns. Spotting good values of $x$ to substitute can reduce the amount of work.

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so it's just a matter of practice to spot the best values for x? Following the guidance of the book they recommended to start with 1. by equating the coefficients you mean place on either side? it would be helpful to give me a step by step breakdown. – peter_gent Jun 11 '13 at 14:29
Starting with $1$ eliminates the terms in $A$ and $B$ so isolates $C$ immediately. Calculations like this occur in methods which involve "partial fractions" for example. There are then standard methods and short-cuts. – Mark Bennet Jun 11 '13 at 14:31

Comparing lead coef's, $\,A = 3.\,$ At $\,x=1\,\Rightarrow\,C = 4,\,$ so at $\,x=-1\,\Rightarrow\,B=2.$

As above, generally for linear factors, a judicious choice of evaluation and coefficient comparisons easily yields the result. This is sometimes called the Heaviside cover up method, esp. when solving equations arising from partial-fraction decompositions. It deserves to be better known that analogous methods extend to nonlinear factors.

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heavyside cover up seems far too advanced for what is required. – peter_gent Jun 11 '13 at 14:51
@peter_gent Huh? The methods used in all the answers are nothing but said method (extracted from the context of partial fractions, where equations of this form arise). If you examine the proof of the Heaviside method then the relationship will be clear, i.e. clear denominators and proceed as above. – Key Ideas Jun 11 '13 at 14:55

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