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Write a polynomial of least degree with real coefficients and with the root 1βˆ’6i1-6i.

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Q. Write a polynomial of least degree with real coefficients and with the root 1βˆ’6i1-6i.
  1. Recognize Property of Complex Roots: Recognize that if a polynomial has real coefficients and a complex root, then the complex conjugate of that root must also be a root of the polynomial. The complex conjugate of 1–6𝑖1–6𝑖 is 1+6𝑖1+6𝑖.
  2. Write Root Factors: Write the factors associated with the roots 1βˆ’6𝑖1-6𝑖 and 1+6𝑖1+6𝑖. The factors are (xβˆ’(1βˆ’6𝑖))(x - (1 - 6𝑖)) and (xβˆ’(1+6𝑖))(x - (1 + 6𝑖)).
  3. Multiply Factors: Multiply the two factors to obtain the polynomial. We have: \newline(xβˆ’(1βˆ’6𝑖))(xβˆ’(1+6𝑖))=(xβˆ’1+6𝑖)(xβˆ’1βˆ’6𝑖)(x - (1 - 6𝑖))(x - (1 + 6𝑖)) = (x - 1 + 6𝑖)(x - 1 - 6𝑖).
  4. Expand Using FOIL Method: Use the FOIL method to expand the product of the two binomials:\newline(xβˆ’1+6i)(xβˆ’1βˆ’6i)=x2βˆ’xβˆ’6ixβˆ’x+1+6iβˆ’6ix+6i+36i2(x - 1 + 6\mathit{i})(x - 1 - 6\mathit{i}) = x^2 - x - 6\mathit{i}x - x + 1 + 6\mathit{i} - 6\mathit{i}x + 6\mathit{i} + 36\mathit{i}^2.
  5. Simplify Expression: Simplify the expression by combining like terms and using the fact that i2=βˆ’1i^2 = -1: x2βˆ’xβˆ’6ixβˆ’x+1+6iβˆ’6ix+6iβˆ’36=x2βˆ’2x+1βˆ’36=x2βˆ’2xβˆ’35x^2 - x - 6ix - x + 1 + 6i - 6ix + 6i - 36 = x^2 - 2x + 1 - 36 = x^2 - 2x - 35.
  6. Verify Standard Form: Verify that the polynomial is in standard form with a leading coefficient of 11: The polynomial x2βˆ’2xβˆ’35x^2 - 2x - 35 is in standard form with a leading coefficient of 11.

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