Try 2 Bash Me
Find the number real x satisfying the equation below.
x 1 5 − 1 5 x 1 4 + 1 0 5 x 1 3 − 4 5 5 x 1 2 + 1 3 6 5 x 1 1 − 3 0 0 3 x 1 0 + 5 0 0 5 x 9 − 6 4 3 5 x 8 + 6 4 3 5 x 7 − 5 0 0 5 x 6 + 3 0 0 3 x 5 − 1 3 6 5 x 4 + 4 5 5 x 3 − 1 0 5 x 2 + 1 5 x − 1 x 1 6 − 1 6 x 1 5 + 1 2 0 x 1 4 − 5 6 0 x 1 3 + 1 8 2 0 x 1 2 − 4 3 6 8 x 1 1 + 8 0 0 8 x 1 0 − 1 1 4 4 0 x 9 + 1 2 8 7 0 x 8 − 1 1 4 4 0 x 7 + 8 0 0 8 x 6 − 4 3 6 8 x 5 + 1 8 2 0 x 4 − 5 6 0 x 3 + 1 2 0 x 2 − 1 6 x + 1 = 0
The answer is 0.
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1 solution
A hint on how the expressions came out to be ( x − 1 ) 1 5 and ( x − 1 ) 1 6 : Check the Pascal’s triangle and the first two and the last 2 coefficients in each polynomial.
The top is ( x − 1 ) 1 6 , and the bottom is ( x − 1 ) 1 5 . The only solution (when simplified) is 1, however, 1 does not work because the equation would equal 0/0, which is undefined, so the answer is 0 solutions.