+6 P(X) is a polynomial with real coefficients such that P(2+i)=4-3i. Find P(2-i).
+128456 I'm sure that there is a general way to solve this....but ...let the polynomial be
-3x^2 + 9x - 5
So
P(2 + i) =
-3 ( 2 + i)^2 + 9 (2 + i) - 5 =
-3 (4 + 4i + i^2) + 9(2 + i) - 5 =
-3 ( 3 + 4i) + 18 + 9i - 5 =
-9 - 12i + 18 + 9i - 5 =
4 - 3i
So
P ( 2 -i) =
-3 (2 - i)^2 + 9(2 - i) - 5 =
-3 (4 - 4i + i^2) + 18 - 9i - 5 =
-3 ( 3 - 4i) + 18 - 9i - 5 =
-9 + 12i + 18 - 9i - 5 =
4 + 3i
+26367 \(P(x)\) is a polynomial with real coefficients such that \(P(2+i)=4-3i\). Find \(P(2-i)\).
\(\begin{array}{|rcll|} \hline \mathbf{P(x)} &=& \mathbf{ax+b} \\\\ P(2+i) &=& a(2+i)+b \\ P(2+i) &=& 2a+ai+b \\ P(2+i) &=& 2a+b+ai \quad | \quad P(2+i) = 4-3i \\ 4-3i &=& \underbrace{2a+b}_{=4}+\underbrace{a}_{=-3}i \qquad \text{compare} \\\\ \mathbf{a} &=& -3 \\\\ 2a+b &=& 4 \\ 2(-3)+b &=& 4 \\ -6+b &=& 4 \\ \mathbf{b} &=& 10 \\\\ \mathbf{P(x)} &=& \mathbf{-3x+10} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline \mathbf{P(x)} &=& \mathbf{-3x+10} \\ P(2-i) &=& -3(2-i)+10 \\ P(2-i) &=& -6+3i +10 \\ \mathbf{P(2-i)} &=& \mathbf{4+3i} \\ \hline \end{array}\)
