Partial fractions - Repeated linear factors (summary example) : ExamSolutions Maths Revision

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[@dikshantgaba5861]

excellent video ......helped me a lot 😆

[@kspades2530]

Great videos thanks

[@mscomedy624]

THANKS FOR THE SUPPORT KEEP DOING SO

[@sambernstein2952]

Thank you

[@rosep4424]

How to assign correct value/ equation to correct constant when writing them out ?

[@sadmanzaid420]

When A(x-3) + B(x-2) we would want to have *ONLY* *ONE* constant so in order to have a single constant one has to be removed, so this example only _applies_ to this specific context mind you! If we desire one constant let's choose A to keep and B to remove, B(x-2) solve the inside factor of (x-2) by equating it to 0
∴ x-2 = 0
∴ x - 2 + 2 = + 2
∴ x = 2
see that after we solved the factor of x by equating it to zero we can use x = 2 as the correct value to input, when x=2 , A(2-3) + B(2-2)
∴ A(-1) + B(0)
∴ -A Voilà

[@moyadaj]

@@sadmanzaid420 this world needs more people like you.

[@9clpc858]

Why do you need to compare coefficients of x^2?

[@ExamSolutions_Maths]

+9clpc8 You don't have to but it made the calculation easier. Try comparing the x terms or constants or even putting in another value for x. Compare your methods and you should arrive at the same value for B but I think you will agree that comparing coefficients of x^2 was the easiest.

[@sadmanzaid420]

This is an alternative method of solving for constants, this method helps in a myriad of ways as well so its only use isn't limited to partial-fraction constant.