Chapter 2, Problem 22T

Solution

To write: A polynomial function $f$ of least degree that has rational coefficients, a leading coefficient of $1$ and the given zeros.

The required polynomial function would be $f\left(x\right)=x^4-10x^3+32x^2-92x+60$ .

Given information:

Zeros of a function: $1+3i,4+\sqrt{10}$ .

Formula used:

1. Factor Theorem: If $x=a,b,c$ are zeros of the function $f\left(x\right)$ , then $\left(x-a\right),\left(x-b\right)\text{and }\left(x-c\right)$ would be factors of the function.
2. Complex Conjugate Theorem: If $f$ is a polynomial function with real coefficients, and $\left(a+bi\right)$ is an imaginary zero of $f$ , then $\left(a-bi\right)$ is also a zero of $f$ .
3. Irrational conjugates Theorem: Suppose $f$ is a polynomial function with real coefficients, and $a\text{and }b$ are rational numbers such that $\sqrt{b}$ is irrational. If $a+\sqrt{b}$ is a zero of $f$ , then $a+\sqrt{b}$ is also a zero of $f$ .

Calculation:

For the given problem, there are $2$ zeros. As per irrational conjugate theorem and complex conjugate theorem, the other zeros would be $1-3i,4-\sqrt{10}$ .

Using the factor theorem, the factored form of $f$

would be:

  $\begin{array}{c}f\left(x\right)=\left(x-\left(1-3i\right)\right)\left(x-\left(1+3i\right)\right)\left(x-\left(4+\sqrt{10}\right)\right)\left(x-\left(4-\sqrt{10}\right)\right)\\ =\left(x-1+3i\right)\left(x-1-3i\right)\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)\end{array}$

Now expand the factors using the difference of squares $\left(a-b\right)\left(a+b\right)=a^2-b^2$:

  $\begin{array}{c}f\left(x\right)=\left(x-1+3i\right)\left(x-1-3i\right)\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)\\ =\left({\left(x-1\right)}^2-{\left(3i\right)}^2\right)\left({\left(x-4\right)}^2-{\left(\sqrt{10}\right)}^2\right)\\ =\left({\left(x-1\right)}^2-9i^2\right)\left({\left(x-4\right)}^2-10\right)\\ =\left({\left(x-1\right)}^2+9\right)\left({\left(x-4\right)}^2-10\right)\left(\text{Using identity }-i^2=1\right)\\ =\left(x^2-2x+1+9\right)\left(x^2-8x+16-10\right)\left(\text{Using rule }{\left(a-b\right)}^2=a^2-2ab+b^2\right)\\ =\left(x^2-2x+10\right)\left(x^2-8x+6\right)\\ =x^2\left(x^2-8x+6\right)-2x\left(x^2-8x+6\right)+10\left(x^2-8x+6\right)\left(\text{Using distributive property}\right)\\ =x^4-8x^3+6x^2-2x^3+16x^2-12x+10x^2-80x+60\\ =x^4-10x^3+32x^2-92x+60\end{array}$

Therefore, the required function would be $f\left(x\right)=x^4-10x^3+32x^2-92x+60$ .