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# Suppose $P(x)=ax^2+bx+c$ and $P(1)=1,P(2)=2$, and $P(3)=4$. Find $a - b - c$.

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Suppose P(x)=ax^2+bx+c and P(1)=1,P(2)=2, and P(3)=4. Find a - b - c.

Thank you

Aug 29, 2022

#1
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Substituting the info gives us a system with 3 equations:

$$1 = a + b + c$$

$$2 = 4a + 2b + c$$

$$3 = 16a + 4b + c$$

Subtracting the first equation from the second and the second from the third gives us another system:

$$3a + b = 1$$

$$12a + 2b = 1$$

Solving this system gives us $$a = -{1 \over 6}$$ and $$b = 1 {1 \over 2}$$

Substituting this into the first equation gives us $$c = -{1 \over 3}$$, so $$a - b - c = -{1 \over 6} - {3 \over 2} + {1 \over 3} = \color{brown}\boxed{-1 {1 \over 3}}$$

Aug 30, 2022
#2
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That seems to be wrong....

Aug 31, 2022
#3
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We translate the given information about P into information about a, b and c.

\begin{align*}
P(1) &= 1 \implies a+b+c =1,\\
P(2) &= 2 \implies 4a+2b+c =2,\\
P(3) &= 4 \implies 9a+3b+c =4.
\end{align*}

Subtracting the first equation from the second and the second equation from the third gives

\begin{align*}3a+b&=1,\\
5a+b&=2.
\end{align*}

Subtracting the first equation above from the second gives 2a=1, so a=1/2 and b=-1/2. It then follows that c=1. So, our answer is\

$\frac{1}{2} - \left(-\frac 12\right) - 1 = \boxed{0}.$

Aug 31, 2022