Problem 35 - Precalculus Mathematics for Calculus

Question

Find a polynomial with integer coefficients that satisfies the given conditions. \(P\) has degree 2 and zeros \(1+i\) and \(1-i\).

Step-by-Step Solution

Answer

The polynomial is \(x^2 - 2x + 2\).

Step 1: Understand the Conditions

We are asked to find a quadratic polynomial, which means it will have a degree of 2, with integer coefficients. The given zeros of the polynomial are complex conjugates: \(1+i\) and \(1-i\).

Step 2: Recall the Polynomial from Zeros

If \(a\) and \(b\) are the zeros of a polynomial, then the polynomial can be expressed as \(P(x) = (x-a)(x-b)\). In our case, \(a = 1+i\) and \(b = 1-i\).

Step 3: Set Up the Polynomial Expression

Substitute the known zeros into the polynomial expression: \(P(x) = (x - (1+i))(x - (1-i))\).

Step 4: Expand the Expression

Expand \((x - (1+i))(x - (1-i))\). Start by combining like terms:\[(x - 1 - i)(x - 1 + i) = ((x-1) - i)((x-1) + i)\]

Step 5: Apply the Difference of Squares

Recognize \((u-v)(u+v) = u^2 - v^2\) as the difference of squares pattern. Here, \(u = x-1\) and \(v = i\):\[(x-1)^2 - i^2 = (x-1)^2 - (-1) = (x-1)^2 + 1\]

Step 6: Simplify

Now expand \((x-1)^2 + 1\):\[(x-1)^2 + 1 = x^2 - 2x + 1 + 1 = x^2 - 2x + 2\]

Step 7: Confirm Integer Coefficients

The polynomial \(P(x) = x^2 - 2x + 2\) has integer coefficients, and it meets all the given conditions.

Key Concepts

Complex Conjugates

The concept of complex conjugates is crucial when working with quadratic polynomials having complex roots. In simple terms, complex conjugates are pairs of complex numbers. One is the mirror image of the other across the real axis. For a complex number of the form \(a + bi\), its complex conjugate is \(a - bi\).

• For the zeros \(1+i\) and \(1-i\) as given in the polynomial, notice that they are complex conjugates of each other.
• When a polynomial has complex coefficients, complex roots always appear in conjugate pairs. This is due to the fundamental theorem of algebra.

This pairing ensures that the coefficients of the polynomial, when expanded, remain real numbers. This is essential for polynomials with integer coefficients, like in this exercise.

Polynomial Coefficients

Polynomial coefficients are the numbers that multiply the variables or powers in a polynomial. These coefficients play a major role in defining the characteristics of a polynomial. In our case, you want the coefficients to be integers.

• The given polynomial is quadratic, meaning its highest degree term is \(x^2\). This involves three coefficients: one each for \(x^2\), \(x\), and the constant term.
• When the polynomial is expanded from its factors \((x - (1+i))(x - (1-i))\), the coefficients turn out to be integers: specifically, \(1\), \(-2\), and \(2\).

These integer coefficients are derived from the fact that any operations (such as addition, subtraction) upon the terms will result in integer values. This is especially true when the complex parts cancel out, as they do here due to complex conjugates.

Polynomial Expansion

Polynomial expansion is the process of expressing a polynomial from its factored form into a standard polynomial form where no products appear. To find the expanded form, you multiply the factors together, ensuring every term is simplified to contain no parenthetical terms.

• In the exercise, we start with the roots \(1+i\) and \(1-i\). Writing these as factors gives us \((x - (1+i))(x - (1-i))\).
• Using the difference of squares formula, \((a-b)(a+b) = a^2 - b^2\), simplifies the expansion. Here \(a = x-1\) and \(b = i\), leading to \((x-1)^2 - i^2\).

Careful expansion of this expression will yield \(x^2 - 2x + 2\). The difference of squares is instrumental because it uses the conjugate's property to eliminate imaginary parts, leaving you with a polynomial with real, integer coefficients.