If (x-6) is a factor of …

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Posted by Abhishek Raj 8 years, 5 months ago
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Dharmendra Kumar 8 years, 5 months ago
Since (x-6) is a factor of x3+ax2+bx-b=0------(1)
Therefore x=6 is one of the roots of the given cubic polynomial.
So put x=6 in eq.(1) we get,
63+a×62+b×6-b=0
216+36a+6b-b=0
36a+5b=-216 -------(2)
given that a-b=7
a=7+b -----(3)
Substituting the value of a from eq.(3) into eq.(2) we get
36(7+b)+5b=-216
252+36b+5b=-216
41b=-216-252
41b=-468
{tex}b = {-468 \over 41}{/tex}
Substituting this value in eq.(3) we get
a=7+({tex}{-468\over 41}){/tex}={tex}{41×7-468\over 41}{/tex}={tex}{287-468 \over 41}{/tex}={tex}{-181\over 41}{/tex}
4Thank You