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When a polynomial $P(x)$ is divided by $x^2 - 3x$, the quotient and remainder are $Q(x)$ and $2x-8$ respectively. The remainder of division of $Q(x)$ by $x-4$ is $2$. Evaluate the remainder of division of $P(x)$ by $x^2 - 4x$.

From linear division of polynomials, one can conclude that

$$Q(4) = 2$$

I also obtained that

$$P(3) = Q(x)(2x-8)$$

This is where I'm stuck. Could you assist me?

Regards

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  • 3
    $\begingroup$ I will translate the conditions into equations. You have $Q(x)=R(x)(x-4)+2$ for some $R(x)$ unknown polynomial. You have $P(x)=(x^2-3x)Q(x)+2x-8$. Can you solve it from here? $\endgroup$ – Alec B-G Feb 6 at 16:21
  • $\begingroup$ Anyone cannot help with that? $\endgroup$ – Enzo Feb 6 at 21:58
  • $\begingroup$ I'll write a bit more $P(x)=x(x-4)(x-3)R(x)+2(x^2-3x)+2x-8$. $\endgroup$ – Alec B-G Feb 7 at 12:37

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