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Let

$$P(x)=x^4+ax^3+bx^2+cx+d$$

where $a,b,c$ and $d$ are constants.

If $P(1)=10,P(2)=20$ and $P(3)=30$, then what is the value of $$\frac{P(12)+P(-8)}{10}$$

I tried substituting $x=1,2,3$ in $P(x)$ to get three linear equations in $a,b,c$ and $d$, but it ended up being quite difficult. $$a+b+c+d=9$$ $$8a+4b+2c+d=4$$ $$27a+9b+3c+d=-51$$

Is there a simpler method to this question?

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  • $\begingroup$ Solve for $a,b,c$ in terms of $d$ $\endgroup$ Apr 2, 2018 at 5:27
  • $\begingroup$ Out of curiosity, what year is this question from? $\endgroup$ Apr 2, 2018 at 14:48
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    $\begingroup$ @JoonasIlmavirta I did not get you, this question was asked in a test I wrote recently $\endgroup$
    – Fasal123
    Apr 2, 2018 at 14:49
  • $\begingroup$ @Fasal123 The answer is a year from a couple of decades ago. It can be a coincidence, though. It is quite common in contest mathematics to engineer questions so that the answer is the current year. I was just trying to figure out whether that had happened here. $\endgroup$ Apr 2, 2018 at 15:01
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    $\begingroup$ @JoonasIlmavirta No, I just think it is a coincidence $\endgroup$
    – Fasal123
    Apr 2, 2018 at 15:03

2 Answers 2

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Define $Q(x) = P(x) - 10x$, then$$ P(1) = 10,\ P(2) = 20,\ P(3) = 30 \Longrightarrow Q(1) = Q(2) = Q(3) = 0. $$ Since $\deg Q = 4$ and $Q$ is monic, there exists $x_0 \in \mathbb{R}$ such that$$ Q(x) = (x - 1)(x - 2)(x - 3)(x - x_0). $$ Therefore,\begin{align*} &\mathrel{\phantom{=}}{} P(12) + P(-8)\\ &= Q(12) + Q(-8) + 40\\ &= 11 × 10 × 9 × (12 - x_0) + (-9) × (-10) × (-11) × (-8 - x_0) +40\\ &= 11 × 10 × 9 × (12 - x_0) + 11 × 10 × 9 × (8 + x_0) +40\\ &= 11 × 10 × 9 × ((12 - x_0) + (8 + x_0)) +40\\ &= 11 × 10 × 9 × 20 +40\\ &=19840. \end{align*}

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$$P(x) = (x-1)(x-2)(x-3)(x-\alpha) + 10 x$$

$$P(12) = 990(12- \alpha) + 120 \tag1$$

$$P(-8) = -990(-8 - \alpha) - 80 \tag 2$$

Adding $(1)$ and $(2)$,
$$P(12) + P(-8) = 19840 \implies \cfrac{P(12) + P(-8)}{10} = 1984$$

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