Determine the real coefficients b and c such that the…

Determine the real coefficients $b$ and $c$ such that the polynomial $P(x) = x^2+bx+c$ satisfies the condition $P(1-i) = 2+i$.

Step 1 · Compute $(1-i)^2$

$$ (1-i)^2 = 1-2i+i^2 = 1-2i-1 = -2i $$

Step 2 · Evaluate $P(1-i)$

$$ P(1-i) = (1-i)^2+b(1-i)+c = -2i+b-bi+c = (b+c)+i(-2-b) $$

Step 3 · Set up the equation

$$ (b+c)+i(-2-b) = 2+i $$

Step 4 · Equate real and imaginary parts

$$ \text{Real: } b+c=2 \qquad\qquad \text{Imaginary: } -2-b=1 $$

Step 5 · Solve

From the imaginary part: $-2-b=1 \Rightarrow -b=3 \Rightarrow b=-3$.

Substituting into the real part: $-3+c=2 \Rightarrow c=5$.

$$ \boxed{b=-3, \quad c=5} \qquad\Longrightarrow\qquad P(x) = x^2-3x+5 $$

Step 6 · Verify

$$ P(1-i) = (1-i)^2-3(1-i)+5 = -2i-3+3i+5 = (-3+5)+i(-2+3) = 2+i \quad\checkmark $$

Step 7 · Cross-check via the conjugate

Since $P$ has real coefficients, $P(\bar z)=\overline{P(z)}$, so $P(1+i)$ should equal $\overline{2+i}=2-i$. Computing $(1+i)^2=1+2i+i^2=2i$:

$$ P(1+i) = 2i-3(1+i)+5 = 2i-3-3i+5 = (-3+5)+i(2-3) = 2-i \quad\checkmark $$

confirming the symmetry property and giving an independent check on $b,c$.

Step 8 · As a linear system

The two conditions can be written as a $2\times2$ real linear system in $b,c$:

$$ \begin{cases} b+c=2 \\ -b\ \ \ \ \,=3 \end{cases} \qquad\Longleftrightarrow\qquad \begin{pmatrix}1&1\\-1&0\end{pmatrix}\begin{pmatrix}b\\c\end{pmatrix}=\begin{pmatrix}2\\3\end{pmatrix} $$

immediately giving $b=-3$ from the second row, then $c=5$ from the first — matching Step 5.

Step 9 · A look at the resulting polynomial

The roots of $P(x)=x^2-3x+5$ are:

$$ x = \frac{3\pm\sqrt{9-20}}{2} = \frac{3\pm i\sqrt{11}}{2} $$

a complex-conjugate pair, since the discriminant $9-20=-11<0$. The vertex of the parabola sits at $x=\dfrac32$, where $P\left(\dfrac32\right) = \dfrac94-\dfrac92+5=\dfrac{11}{4}>0$ — consistent with the graph never touching the real axis.