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# Can someone drive me in the right direction or help me? Please put solution. ty :D

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P(X) is a polynomial with real coefficients such that P(2+i)=4-3i. Find P(2-i).

Jan 9, 2020

#1
+107355
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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

Jan 9, 2020
#2
+24093
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$$P(x)$$ is a polynomial with real coefficients such that $$P(2+i)=4-3i$$. Find $$P(2-i)$$.

$$\begin{array}{|rcll|} \hline \mathbf{P(x)} &=& \mathbf{ax+b} \\\\ P(2+i) &=& a(2+i)+b \\ P(2+i) &=& 2a+ai+b \\ P(2+i) &=& 2a+b+ai \quad | \quad P(2+i) = 4-3i \\ 4-3i &=& \underbrace{2a+b}_{=4}+\underbrace{a}_{=-3}i \qquad \text{compare} \\\\ \mathbf{a} &=& -3 \\\\ 2a+b &=& 4 \\ 2(-3)+b &=& 4 \\ -6+b &=& 4 \\ \mathbf{b} &=& 10 \\\\ \mathbf{P(x)} &=& \mathbf{-3x+10} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{P(x)} &=& \mathbf{-3x+10} \\ P(2-i) &=& -3(2-i)+10 \\ P(2-i) &=& -6+3i +10 \\ \mathbf{P(2-i)} &=& \mathbf{4+3i} \\ \hline \end{array}$$

Jan 9, 2020
#3
+107355
+1

Thanks, heureka.......that is a more general way to derive this....!!!!

CPhill  Jan 9, 2020