1Following Nishant sir's hint .....
(x3 - x) =x(x - 1)(x + 1)
f(x) = x(x + 1)(x - 1)g(x) + ax + b .... (1)
f(x) = xh(x) + 1 ....... (2)
f(x) = (x - 1)m(x) + 3 ....... (3)
f(x) = (x + 1)q(x) + 1 ....... (4)
Now this can be solved .....
Let f(x) = x135 + x125 - x115 + x5 + 1. If f(x) is divided by x3 - x, then the remainder is some function of x say g(x). Find the value of g(10).
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6 Answers
62f(x) on division by x leaves a remainder of 1
f(x) on division by (x-1) leaves a remainder of f(1) = 1+1-1+1+1 = 3
f(x) on division by (x+1) leaves a remainder of f(-1) = -1-1+1-1+1 = 1
g(x) will be a quadratic function.. because it is the remainder on division by a cubic equation.
Now can you try?
162Another method that I can suggest is (I was hoping that someone else can figure it from my hint)
f(x)=g(x)(x-1)(x+3) + a(x-1) + b
where a and b are constants....
Now try to solve...
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- Anonymous Where did i get (x-1), (x+3), a(x-1) + b ?Upvote·0·2018-07-22 12:45:42
341Sometimes there are fun ways of doing these, especially when they invite you with large exponents.
x^{135}+x^{125}-x^{115}+x^5+1 = (x^{135}-x^{115})+(x^{125}-x^5)+ 2(x^5-x)+(2x+1)
Each of the bracketed expressions is divisible by x^3-x leaving the remainder 2x+1 to be read off.