Question 1203631
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{{{A/(x-1)+B/(x-2)+C/(x-3)=(2x^2-6x+6)/((x-1)(x-2)(x-3))}}}


{{{(x-1)(x-2)(x-3)*(A/(x-1)+B/(x-2)+C/(x-3))=(x-1)(x-2)(x-3)*((2x^2-6x+6)/((x-1)(x-2)(x-3)))}}} Multiply both sides by the LCD to clear out fractions 


{{{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)=2x^2-6x+6}}}


{{{A(x^2-5x+6)+B(x^2-4x+3)+C(x^2-3x+2)=2x^2-6x+6}}}


{{{Ax^2-5Ax+6A+Bx^2-4Bx+3B+Cx^2-3Cx+2C=2x^2-6x+6}}}


{{{(Ax^2+Bx^2+Cx^2)+(-5Ax-4Bx-3Cx)+(6A+3B+2C)=2x^2-6x+6}}}


{{{(A+B+C)x^2+(-5A-4B-3C)x+(6A+3B+2C)=2x^2-6x+6}}}


Equate corresponding terms
(A+B+C)x^2 = 2x^2 leads to A+B+C = 2
(-5A-4B-3C)x = -6x leads to -5A-4B-3C = -6
6A+3B+2C = 6


The task is to solve this system of equations
A+B+C = 2
-5A-4B-3C = -6
6A+3B+2C = 6


Use either elimination, substitution, or <a href="https://www.algebra.com/algebra/homework/coordinate/Linear-systems.faq.question.1203611.html">matrix row reduction</a>. 


I'll skip the steps for this part, but the answers should be:
<font color=red>A = 1
B = -2
C = 3</font>


Therefore,
{{{A/(x-1)+B/(x-2)+C/(x-3)=(2x^2-6x+6)/((x-1)(x-2)(x-3))}}}
updates to
{{{1/(x-1)-2/(x-2)+3/(x-3)=(2x^2-6x+6)/((x-1)(x-2)(x-3))}}}
It is an identity that is true for nearly all real numbers x but {{{x <> 1}}} and {{{x <> 2}}} and {{{x <> 3}}} (to avoid division by zero errors).


WolframAlpha can be used to confirm the answer.
GeoGebra's CAS feature can also be used.
There are many other online calculators that offer the same features.


A different approach can be found here:
<a href="https://www.algebra.com/algebra/homework/expressions/expressions.faq.question.1196220.html">https://www.algebra.com/algebra/homework/expressions/expressions.faq.question.1196220.html</a>
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