# What will be the value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$?

What will be the value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$ provided that $$P(1)=10$$ $$P(2)=20$$ $$P(3)=30$$

I put these values and got three simultaneous equations in $a, b, c, d$. What is the smarter way to approach these problems? Please help.

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you should find $a$,$b$,$c$,$d$ by initial conditions and then directly put $12$ and $-8$ –  dato datuashvili Jul 18 '12 at 11:52
Please can you show your few steps?I'm unable to do so.. –  TheApe Jul 18 '12 at 12:01
also i think that ,these equations have infinity solution,just take any value of d and solve equation,you will have three equation with three unknown value –  dato datuashvili Jul 18 '12 at 12:02
54 minutes.  –  Did Jul 18 '12 at 12:35
Sir did, you did it again. –  TheApe Jul 18 '12 at 12:38

Two remarks, to avoid almost every computation:

• The polynomial $P(x)-10x$ has roots $1$, $2$ and $3$, hence there exists a polynomial $Q$ such that $P(x)-10x=(x-1)(x-2)(x-3)Q(x)$.
• The polynomial $P(x)-10x$ has degree $4$ and leading coefficient $1$, hence $Q(x)=x+z$ for some unknown constant $z$ whose value will be irrelevant.

Thus, $P(12)+P(-8)=10\cdot(12-8)+11\cdot10\cdot9\cdot(12+z)+9\cdot10\cdot11\cdot(8-z)$, that is, $P(12)+P(-8)=10\cdot4+11\cdot10\cdot9\cdot(12+z+8-z)=40+990\cdot20=19840$.

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I'm not sure this is the smartest way, but here's one way.

Define $Q(x) = P(x) - x^4$ which satisfies the conditions $$Q(1) = 9, Q(2) = 4, Q(3) = -51.$$

So $a,b,c,d$ satisfy the matrix identity

$$\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \\ 1 & 3 & 9 & 27 \end{pmatrix} \begin{pmatrix} d \\ c \\ b \\ a \end{pmatrix} = \begin{pmatrix} 9 \\ 4 \\ -51 \end{pmatrix}.$$ To solve this, find the kernel of the matrix (a bit of linear algebra), which turns out to be $$<\begin{pmatrix} -6 \\ 11 \\ -6 \\ 1 \end{pmatrix}>$$ and add a particular solution, for example where $a = 0$: $b = -25, c = 70, d = -36$.

So your polynomial is given by $$P(x) = x^4 + ax^3 + (-25-6a)x^2 + (11 + 70a)x + (-36 - 6a)$$ for some $a$ which we can't determine.

You then get $$P(12) + P(-8) = 17940 + 990a + 1900 - 990a = 19840.$$

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Other way doing this:

We try to find reals $e,f,g$ such that $P(12)+P(-8)=eP(1)+fP(2)+gP(3)$. So, if we try to equal "$x^k$ evaluated", we gain a system of equations: $$\left\{\begin{array}{ccc} 1^ke+2^kf+3^kg&=&12^k+(-8)^k \end{array}\right.,\quad k=0,\cdots,4$$ In particular, $$\left\{\begin{array}{ccc} e+f+g&=&1+1\\ e+2f+3g&=&12+(-8)\\ e+2^2f+3^2g&=&12^2+(-8)^2 \end{array}\right.$$ and we obtain $e=100,f=-198,g=100$. Verifying the others values for $k$: $$\left\{\begin{array}{ccc} 100+2^3(-198)+3^3\cdot100&=&12^3+(-8)^3\\ 100+2^4\cdot(-198)+3^4\cdot100&=&12^4+(-8)^4 + 19800 \end{array}\right.$$

So, $P(12)+P(-8)=100P(1)-198P(2)+100P(3)+1980=19840$

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if you put values $1$,$2$ and $3$ ,you will get

$a+b+c+d=9$

$8a+4b+2c+d=20$

$27a+9b+3c+d=-51$ you can take any value of $d$ and solve equation for $a,b,c$

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And how do you prove that any other value for d gives the same answer ? –  ama Jul 23 '12 at 20:59