Evaluate the constant term of $P(x-1)$ if the remainder of the division of $P(x)$ by $x-3$ is $18$ and $P(x+1) = (x^2 -4)Q(x)+3ax+6$

Question

Assume that $$P(x+1) = (x^2 -4)Q(x)+3ax+6$$ and that the remainder of the division of polynomial $P(x)$ by $x-3$ is $18$. Evaluate the constant term of polynomial $P(x-1)$.

All I could see so far is that the polynomial $P(x)$ should be quadratic because $Q(x)$ is multiplied by quadratic term, which is $x^2$.

Answer 1

the remainder of the division of polynomial $P(x)$ by $x−3$ is $18$

implies: $$P(x)=(x-3)R(x)+18 \Rightarrow \\ P(x+1)=(x-2)R(x+1)+18=(x^2 -4)Q(x)+3ax+6\\ P(2+1)=18=6a+6 \Rightarrow a=2.$$ Hence: $$P(\color{red}x-1)=P(x-2+1)=((x-2)^2-4)Q(x-2)+6(x-2)+6 =\\ (x^2-4x)Q(x-2)+6x-6 \Rightarrow \\ P(\color{red}0-1)=-6.$$

Answer 2

We are given that $P(x+1) = (x^2 -4)Q(x)+3ax+6$ $(\ast)$. Plugging $x=2$ into $(\ast)$ yields $P(3)=6a+6$. On the other hand, for every $b$, $P(b)$ is the remainder of the division of polynomial $P(x)$ by $x-b$ hence $P(3)=18$, which shows that $a=2$.

Finally, plugging $x=-2$ into $(\ast)$ yields $P(-1)=-6a+6=-6$ hence the constant term of $P(x-1)$ is $P(-1)=-6$.

Answer 3

Let $F(x)=P(x-1)$. Of course, the constant term of $F(x)$ is $F(0)=P(-1)$

We write $$P(x)=g(x)\times (x-3)+18$$ Then, of course, $P(3)=18$

Using the equation $$P(x+1) = (x^2 -4)Q(x)+3ax+6$$ We set $$x=-2\implies P(-1)=-6a+6$$

and $$x=2\implies P(3)=6a+6$$ Since we already know that $P(3)=18$ we deduce that $a=2$ hence $$P(-1)=-12+6=-6$$

Answer 4

It's worth emphasizing how this boils down to determining a line from a point and its $y$-intercept.

We are given $\,f(x) := P(x+1) = (x\color{#c00}{-2})(x\color{#c00}{+2})Q(x) + y(x),\ \ \ y(0)=6,\ \deg(y) \le 1$

and also that $\,f(2)=P(3)=18,\,$ and we seek $f(-2)= P(-1) = $ constant term of $P(x-1)$

But $\,f(\color{#c00}{\pm 2}) = y(\pm 2),\,$ so equivalently we're given $\,y(2)=18\,$ and $\,y(0)=6\,$ and we seek $\,y(-2)$

But we know since grade-school how to determine a line $\,y(x) = mx+b\,$ given both a point on the line and the $y$-intercept $\,b=y(0)\ $ [ditto given a point and the line's slope $\,m = y'(0) = y'$].

Remark $ $ In an abstract algebra course we learn how to view the above problem as a special case of CRT = Chinese Remainder Theorem, i.e. how to compute $\ y = f\bmod (x\!-\!2)(x\!+\!2)\,$ from $\,f(2) = f\bmod x\!-\!2\,$ and $\,f(-2) = f\bmod (x\!+\!2),\,$ e.g. see here. Also we learn the close relationship with various interpolation methods (Lagrange / Newton).