Chapter 2.5, Problem 58E

To determine

To find : all the zeros of the function

Answer to Problem 58E

The zeros of the functionare: $\boxed{\boxed{x=-1,-4+i,\&-4-i}}$

Explanation of Solution

Given information:

  $Function:g\left(x\right)=x^3+9x^2+25x+17;Zero:-4+i$

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 the polynomial $f\left(x\right)$ , 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 $-4+i$ is a zero, the complex conjugate $-4-i$ is also a zero of a function

If $x=-4+i$ is a zero, then $x+4-i$ is factor of the polynomial

If $x=-4-i$ is a zero, then $x+4+i$ is factor of the polynomial

If $x+4-i$ & $x+4+i$ are factors of polynomial, then its product $\left(x+4-i\right)\left(x+4+i\right)$ is also factor of polynomial.

Expanding $\left(x+4-i\right)\left(x+4+i\right)$ using the algebraic identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$

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

Replacing $i^2$ with -1

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

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

  ${\left(x+4\right)}^2+1=x^2+8x+16+1$

Combine like terms

  $x^2+8x+17$

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

Writing the function as product of factors

  $\begin{array}{l}g\left(x\right)=x^3+9x^2+25x+17\\ =\left(x+1\right)\left(x^2+8x+17\right)\end{array}$

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

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

$x+1=0$

  $x+1-1=0-1$

  $x=-1$

Conclusion:

The zeros of the given polynomial

  $g\left(x\right)=x^3+9x^2+25x+17$ are: $x=-1,-4+i,\&-4-i$