13yup .....the degree should be one less
Ax2+bx+c should be upon (x-1)3
dx+e...upon (x-1)2
SHOULD
(x−1)3(x−2)x3
BE WRITTEN AS
(x−1)3A+(x−1)2B+(x−1)C+(x−2)D
IF NOT WHAT IS THE CORRECT WAY TO SPLIT IT?
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39 Answers
11bUT VISHAL, SHOULDN'T THE DEGREE IN THE UMERATOR BE LESS THAN THE DEGREE OF THE DENOMINATOR?
13
1arey hain main thoda confuse ho gaya tha [4][4][4][4]
k 1 bar fir banata huin
1yes cube ke upar numerator me highest power will be 2...
similarly others..
11Why is the expression in the first post of this thread wrong?
When we can write
(x−1)2(x−2)x2
as
x−1A+(x−1)2B+x−2C
why can't we split the expression i gave in to the form i gave??
1(x−1)3Ax2+Bx+C+(x−1)2Dx+E+(x−1)F+(x−2)G