Chapter 2.5, Problem 57E

To determine

To find : all the zeros of the function

Answer to Problem 57E

The zeros of the functionare: $\boxed{\boxed{x=2,3+2i,\&3-2i}}$

Explanation of Solution

Given information:

  $Function:g\left(x\right)=x^3-8x^2+25x-26;Zero:3+2i$

Concept Involved:

When f is a polynomial function and $a$ is a real number, thefollowing are equivalent:

(i) $x=a$ is a zero of $f$ , (ii) $x=a$ is asolution of the equation $f\left(x\right)=0$ , (iii) $\left(x-a\right)$ is a factor of thepolynomial $f\left(x\right)$ , and (iv) $\left(a,0\right)$ is an x-intercept of the graph of $f$ .

Complex Zeros Occur in Conjugate Pairs: Let f be a polynomial function that has real coefficients. If $a+bi$ , where $b\ne 0$ , is a zero of the function, then the complex conjugates $a-bi$ is alsoa zero of the function.

Calculation:

Given that $3+2i$ is a zero, the complex conjugate $3-2i$ is also a zero of a function

If $x=3+2i$ is a zero, then $x-3-2i$ is factor of the polynomial

If $x=3-2i$ is a zero, then $x-3+2i$ is factor of the polynomial

If $x-3-2i$ & $x-3+2i$ are factors of polynomial, then its product $\left(x-3-2i\right)\left(x-3+2i\right)$ is also factor of polynomial.

Expanding $\left(x-3-2i\right)\left(x-3+2i\right)$ using the algebraic identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$

  ${\left(x-3\right)}^2-{\left(2i\right)}^2={\left(x-3\right)}^2-4i^2$

Replacing $i^2$ with -1

  ${\left(x-3\right)}^2-4\left(-1\right)={\left(x-3\right)}^2+4$

Expanding further using the algebraic identity ${\left(a-b\right)}^2=a^2-2ab+b^2$

  ${\left(x-3\right)}^2+4=x^2-6x+9+4$

Combine like terms

  $x^2-6x+13$

Use the long division to find the other factor of the function

  $\begin{array}{l}x-2\\ x^2-6x+13)\overline{x^3-8x^2+25x-26}\\ \underset{\_}{\underset{(-)}{+}{x}^3\underset{(+)}{-}6x^2\underset{(-)}{+}13x}\\ -2x^2+12x-26\\ \underset{\_}{\underset{(+)}{-}2x^2\underset{(-)}{+}12x\underset{(+)}{-}26}\\ \underset{\_}{0}\end{array}$

Writing the function as product of factors

  $\begin{array}{l}g\left(x\right)=x^3-8x^2+25x-26\\ =\left(x-2\right)\left(x^2-6x+13\right)\end{array}$

To set up the third zero of the polynomial, set the factor $x-2$ to zero and solve for x

• Adding2 to both sides of the equation, and then
• Simplifying on both sides of the equation, and then

$x-2=0$

  $x-2+2=0+2$

  $x=2$

Conclusion:

The zeros of the given polynomial

  $g\left(x\right)=x^3-8x^2+25x-26$ are: $x=2,3+2i,\&3-2i$