Given $P(x)=99(x^{101}-1)-101(x^{99}-1)$, find $Q(1)$ where $Q$ is the quotient of the division between $P(x)$ and $(x-1)^3$.

Obviously I could just divide the polynomials but that is not the solution I want. It is possible to figure out $Q(1)$ without doing long division, and that is the answer I am interested in.

  • 1
    $\begingroup$ I'm not quite clear on the question, actually. $P(x)=(x-1)^3Q(x)+R(x)$ where $R(x)$ is quadratic. To obtain $Q(1)$ differentiate each side three times with respect to $x$. $R$ vanishes and the derivatives of $Q$ get multiplied by zero when you put $x=1$. But is this what is intended? $\endgroup$ – Mark Bennet Aug 28 '17 at 21:18

If we write: $$ P(x) = (x-1)^3Q(x)+ax^2+bx+c$$ then $$P'''(x) =6Q(x)+(x-1)[......]$$ On the other hand we have $$P'''(x)= 99\cdot 101 (100\cdot 99x^{98}-98\cdot 97x^{96})$$ So $$ 6Q(1) = P'''(1) = 99\cdot 101 (100\cdot 99-98\cdot 97)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.