Chapter 2.5, Problem 60E

To determine

To find : all the zeros of the function

Answer to Problem 60E

The zeros of the functionare: $\boxed{\boxed{x=1,2,-2+\sqrt{3}i,\&-2-\sqrt{3}i}}$

Explanation of Solution

Given information:

  $Function:h\left(x\right)=x^4+x^3-3x^2-13x+14;Zero:-2+\sqrt{3}i$

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 $-2+\sqrt{3}i$ is a zero, the complex conjugate $-2-\sqrt{3}i$ is also a zero of a function

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

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

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

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

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

Replacing $i^2$ with -1

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

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

  ${\left(x+2\right)}^2+3=x^2+4x+4+3$

Combine like terms

  $x^2+4x+7$

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

  $x^2-3x+2$

  $x^2+4x+7)\overline{x^4+x^3-3x^2-13x+14}$

  $\underset{\_}{\underset{(-)}{+}{x}^4\underset{(-)}{+}4x^3\underset{(-)}{+}7x^2}$

  $-3x^3-10x^2-13x$

  $\underset{\_}{\underset{(+)}{-}3x^3\underset{(+)}{-}12x^2\underset{(+)}{-}21x}$

  $\underset{\_}{\begin{array}{l}2x^2+8x+14\\ \underset{(-)}{+}2x^2\underset{(-)}{+}8x\underset{(-)}{+}14\end{array}}$

  $\underset{\_}{0}$

Writing the function as product of factors

  $\begin{array}{l}h\left(x\right)=x^4+x^3-3x^2-13x+14\\ =\left(x^2+4x+7\right)\left(x^2-3x+2\right)\end{array}$

To set up the third zero of the polynomial, set the factor

  $x^2-3x+2$ to zero

  $x^2-3x+2=0$

To solve the quadratic equation of the form $a{x}^2+bx+c=0$ we can use factoring method:

1. Identify a, b and c of the equation
$a=1;b=-3;c=2$
2. Rewrite the equation like $a{x}^2+mx+nx+c=0$ where m and n are two numbers whose sum equals b and product equals ac. {List the factor pairs of product and see which two factors add up to give b to get the two numbers m and n}
$x^2-2x-1x+2=0$
3. Factor the greatest common factor of first two terms and greatest common factor of last two terms
$x\left(x-2\right)-1\left(x-2\right)=0$
4. Factor the common binomial from the whole expression in the left side.
$\left(x-2\right)\left(x-1\right)=0$

Applying the Zero Product Property to find the values ofx, the Principle of Zero Products states that if $pq=0$ , then either $p\text{=}0$ or $q\text{=}0$ , or both $p$ and $q$ are 0. Once the polynomial is factored, set each factor equal to zero and solve them separately.

  $x-2=0$  ▸ 1^st equation

  $x-1=0$  ▸ 2^nd equation

5. 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

  $x-1=0$

  $x-1+1=0+1$

  $x=1$

6. Set the factor $x-2$ to zero and solve for x
• Adding 2 to both sides of the equation, and then
• Simplifying on both sides of the equation

  $x-2=0$

  $x-2+2=0+2$

  $x=2$

Conclusion:

The zeros of the given polynomial

  $h\left(x\right)=x^4+x^3-3x^2-13x+14$ are: $x=1,2,-2+\sqrt{3}i,\&-2-\sqrt{3}i$