A question on polynomials and functional equations.

Question

If $P(x)$ is a cubic polynomial which satisfies

• $(x-1) ^2$ is a factor of $P(x) +2 $ and $(x+1) ^2$ is a factor of $P(x)-2$

then what is the value of $P(3)$ ?

Please give a general solution. And also a shorter way for MCQ's

Answer 1

By hypothesis, $$(x-1)^2g(x)-2=P(x)=(x+1)^2h(x)+2$$ for some $h(x)=mx+a$ and $g(x)=mx+b$.

$$x=1 \implies -2=4h(1)+2$$ $$x=-1\implies 2=4g(-1)-2$$

so $h(1) = -1$ and $g(-1) = 1$. Thus $$-1=m+a\\1=-m+b$$ so $a+b=0$

Also $$P(0)=g(0)-2=h(0)+2 \implies b-2=a+2 \implies a-b=-4$$

Solving $a$ and $b$, we get $a=-2$ and $b=2$ and hence $m=1$

Thus $g(x)=x+2$ and $h(x)=x-2$ and hence $$P(x)=(x-1)^2(x+2)-2=(x+1)^2(x-2)+2=x^3-3x$$

So $P(3)=3^3-3^2$

Answer 2

Let $P(x) = Ax^3 + Bx^2+Cx+D$. Then $P'(x) = 3Ax^2 + 2Bx + C$.

For the graph $y=P(x)$, $(1,-2)$ and $(-1, 2)$ are stationary points (and actually turning points).

$$\begin{align*} -2 &= A+B+C+D\\ 0 &= 3A + 2B+C\\ 2 &= -A + B-C+D\\ 0 &= 3A-2B+C \end{align*}$$

Solve for $3^3A + 3^2B+3C + D$.

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$(x-1)^2$ is a factor of $P(x)+2$, i.e. $x=1$ is a double (or higher-order) root of $P(x)+2$. So cubic polynomial $P(x)$ has a stationary point at $x=1$, and similarly another stationary point at $x=-1$.

I did not think a lot about it when I first answered, just scribbled a cubic curve in my head. But if you want a proof: let $P(x)+2 = (x-1)^2g(x)$, where $g(x)$ is a linear polynomial. Then by differentiating both sides,

$$\begin{align*} \frac d{dx}(P(x)+2) &= \frac d{dx} \left[(x-1)^2g(x)\right]\\ P'(x) &= (x-1)^2g'(x) + g(x)\cdot 2(x-1)\\ &= (x-1)\left[(x-1)g'(x) + 2g(x)\right]\\ P'(1) &= 0 \end{align*}$$

i.e. $P(x) +2$ and therefore $P(x)$ have a stationary point at $x=1$.