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Let $p(x)=x^4+ax^3+bx^2+cx+d$ where a,b,c,d are constants. If $p(1)=10$, $p(2)=20$, $p(3)=30$, compute $\frac {p(12)+p(-8)}{10}$. I have tried so far. \begin{align} a+b+c+d=&9\\8a+4b+2c+d=&4\\27a+9b+3c+d=&-51 \end{align} Manipulating these, I got $6a+b=-25$. Now, $$\frac {p(12)+p(-8)}{10}=\frac{24832+1216a+208b+4c+2d}{10}$$ $$=\frac{24832+202(6a+b)+(4a+4b+4c)+2b+2d}{10}$$ $$=\frac{19782+(36-4d)+2b+2d}{10}$$ $$=\frac{19818+2b-2d}{10}$$ How do I get rid of the $2b-2d$?

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  • $\begingroup$ Incidentally, the term for a degree $4$ polynomial is "quartic." $\endgroup$ – Cameron Buie Nov 1 '13 at 14:23
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    $\begingroup$ You made a slight error. You should have $2d$ in the numerator, and not $d$. $\endgroup$ – Cameron Buie Nov 1 '13 at 14:29
  • $\begingroup$ And I'm pretty sure that a biquadratic polynomial means a composition of two quadratic polynomials, that is $p(x)=q_1(q_2(x))$ where $q_1,q_2$ are quadratic. $\endgroup$ – tomasz Nov 20 '13 at 18:12
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You have $p(x) = x^4 + ax^3 + bx^2 + cx + d$, and you're given that $a + b + c + d = 9$, $8a + 4b + 2c + d = 4$, and $27a + 9b + 3c + d = -51$

Now, $p(12) + p(-8) = 12^4 + 8^4 + (12^3 - 8^3) a + (12^2 + 8^2) b + (12 - 8) c + 2d = 24832 + 1216 a + 208 b + 4c + 2d = 24832 + 1216 a + 208 b + 2 (2c+d)$. You note that $2c + d = 4 - 8a - 4b$, and substitute that into the equation to get $p(12) + p(-8) = 24832 + 1216 a + 208 b + 8 - 16a - 8b = 24840 + 1200 a + 200b = 24840 + 200(6a + b).$

Plug in the $6a + b = -25$ and you get $p(12) + p(-8) = 24840 - 5000 = 19840$. Divide it by $10$ and you get $\displaystyle \frac{p(12) + p(-8)}{10} = 1984$.

Remember, $d + d \ne d$.

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  • $\begingroup$ Doing things differently,I found 1984 ! $\endgroup$ – Claude Leibovici Nov 1 '13 at 14:57
  • $\begingroup$ @ClaudeLeibovici - I just assumed what Tejas found (i.e. with $6a + b = -30$) was correct. Is that wrong? $\endgroup$ – 2012ssohn Nov 1 '13 at 15:01
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    $\begingroup$ I find that 6 a + b = -25. But I started the problem from scratch. Then, there is probably an error in Teja's calculations. $\endgroup$ – Claude Leibovici Nov 1 '13 at 15:16
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    $\begingroup$ I just checked and you are indeed correct. $6a + b = -25$. I'll go ahead and fix that in my answer. $\endgroup$ – 2012ssohn Nov 1 '13 at 15:18
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(2012's answer is correct, but the algebraic manipulations doesn't reveal what is happening. This answer explains why we could calculate the expression, despite not having enough information.)

By the remainder factor theorem, since $p(x) - 10x$ is a monic quartic polynomial with roots 1, 2, and 3, hence

$$p(x) - 10x = ( x-1) (x-2) ( x-3) (x-k), $$

where $k$ is some constant.

Hence, $ p(12) - 120 = 11 \times 10 \times 9 \times (12-k)$ and $p(-8) - (-80) = (-9) \times (-10) \times (-11) \times (-8-k)$.

Note that the coefficients of $ (12-k) $ and $-(-8-k)$ are the same, namely $11\times 10 \times 9$, so we can add them up to get:

$$p(12) + p(-8) = 120 + (-80) + 11 \times 10 \times 9 \times (12+8) = 19840.$$

If you want a similar problem to practice what you learnt in this problem, try this math problem on Brilliant.

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As you noticed, you have three equations for four unknowns (a,b,c,d). Eliminate a, b and c expressing then as functions of d. Then compute your expression. By magics, d disappears ! Does this help ?

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