Chapter 2.5, Problem 55E

To determine

To find : all the zeros of the function

Answer to Problem 55E

The zeros of the functionare: $\boxed{\boxed{x=1,-2i,\&2i}}$

Explanation of Solution

Given information: $Function:f\left(x\right)=x^3-x^2+4x-4;Zero:2i$

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 $2i$ is a zero, the complex conjugate $-2i$ is also a zero of a function

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

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

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

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

  $x^2-{\left(2i\right)}^2=x^2-4i^2$

Replacing $i^2$ with -1

  $x^2-4\left(-1\right)=x^2+4$

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

  $\begin{array}{l}x-1\\ x^2+4)\overline{x^3-1x^2+4x-4}\\ \underset{\_}{\underset{(-)}{+}{x}^3\underset{(-)}{+}0x^2\underset{(-)}{+}4x}\\ -1x^2+0x-4\\ \underset{\_}{\underset{(+)}{-}1x^2\underset{(-)}{+}0x\underset{(+)}{-}4}\\ \underset{\_}{0}\end{array}$

Writing the function as product of factors

  $\begin{array}{l}f\left(x\right)=x^3-x^2+4x-4\\ =\left(x-1\right)\left(x^2+4\right)\end{array}$

To set up the third zero of the polynomial, set the factor $x-1$ to zero and solve for x

• Adding 1 to both sides of the equation, and then
• Simplifying on both sides of the equation

  $x-1=0$

  $x-1+1=0+1$

  $x=1$

Conclusion:

The zeros of the given polynomial $f\left(x\right)=x^3-x^2+4x-4$ are: $x=1,-2i,\&2i$