Chapter 2.5, Problem 65E

To determine

To write : the polynomial as the product of linear factors and list all the zeros of function

Answer to Problem 65E

The function as product of linear factors: $\boxed{\boxed{f\left(x\right)=\left(x-2\right)\left(x+2\right)\left(x-2i\right)\left(x+2i\right)}}$

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

Explanation of Solution

Given information: $f\left(x\right)=x^4-16$

Concept Involved:

Linear Factorization Theorem:If $f\left(x\right)$ is a polynomial of degree $n$ , where $n>0$ , then $f\left(x\right)$ has precisely $n$ linearfactors $f\left(x\right)=a_n\left(x-c_1\right)\left(x-c_2\right)...\left(x-c_n\right)$ where $c_1,c_2,...,c_n$ are complex numbers.

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:

To find the zeros of the function we need to set the function to zero $f\left(x\right)=0$

  $x^4-16=0$

Rewrite the left side of the expression in the form $a^2-b^2$

  ${\left(x^2\right)}^2-4^2=0$

Use the algebraic identity $a^2-b^2=\left(a+b\right)\left(a-b\right)$ to factor the left side of the equation

  $\left(x^2+4\right)\left(x^2-4\right)=0$

Rewrite the factor $x^2-4$ in the left side of the equation in the form $a^2-b^2$

  $\left(x^2+4\right)\left(x^2-2^2\right)=0$

Use the algebraic identity $a^2-b^2=\left(a+b\right)\left(a-b\right)$ to factor $x^2-2^2$ further the left side of the equation

  $\left(x^2+4\right)\left(x-2\right)\left(x+2\right)=0$

We need to set each factor equal to zero using the zero factor property which states that If ab = 0, then a = 0 or b = 0.

  $x^2+4=0$  ▸1^st equation

  $x-2=0$  ▸2^ndequation

  $x+2=0$  ▸3^rd equation

Solve the 1st equation $x^2+4=0$

• By subtracting 4 on both sides, then
• By Splitting the number inside the square root in the right side of the equation, then
• By Taking square root on both sides, then
• By Simplifying square root on both sides of the equation, and then
• By Replacing $i$ for $\sqrt{-1}$

  $x^2+4-4=0-4$

  $x^2=-4$

  $\sqrt{x^2}=\sqrt{-4}$

  $\sqrt{x^2}=\sqrt{-1}\cdot \sqrt{4}$

  $x=\sqrt{-1}\cdot \pm 2$

  $x=\pm 2i$

Solving the 2nd equation $x-2=0$

• By adding 2 on both sides
• By simplifying the equation on both sides

  $x-2+2=0+2$

  $x=2$

Solving the 2nd equation $x+2=0$

• By subtracting 2 on both sides
• By simplifying the equation on both sides

  $x+2-2=0-2$

  $x=-2$

List the zeros of the functions given:

  $x=2i;x=-2i;x=2;x=-2$

If $x=k$ is zero, then $x-k$ will be the factor of the function.

So $x-2,x+2,x-2i$ & $x+2i$ are the factors of the function

Write the function $f\left(x\right)=x^4-16$ as product of linear factors:

  $f\left(x\right)=\left(x-2\right)\left(x+2\right)\left(x-2i\right)\left(x+2i\right)$

Conclusion:

The zeros of the given function $f\left(x\right)=x^4-16$ are: $x=2,-2,2i,\&-2i$

The function written as product of linear factors: $f\left(x\right)=\left(x-2\right)\left(x+2\right)\left(x-2i\right)\left(x+2i\right)$