Chapter 2.5, Problem 66E

To determine

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

Answer to Problem 66E

The function as product of linear factors: $\boxed{\boxed{f\left(y\right)=\left(y-4\right)\left(y+4\right)\left(y-4i\right)\left(y+4i\right)}}$

The zeros of the function are: $\boxed{\boxed{y=4,-4,4i,\&-4i}}$

Explanation of Solution

Given information: $f\left(y\right)=y^4-256$

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 $f\left(y\right)=0$

  $y^4-256=0$

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

  ${\left(y^2\right)}^2-{16}^2=0$

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

  $\left(y^2+16\right)\left(y^2-16\right)=0$

Rewrite the factor $y^2-16$ in the left side of the equation in the form $a^2-b^2$

  $\left(y^2+16\right)\left(y^2-4^2\right)=0$

Use the algebraic identity $a^2-b^2=\left(a+b\right)\left(a-b\right)$ to factor $y^2-4^2$ further the left side of the equation

  $\left(y^2+16\right)\left(y-4\right)\left(y+4\right)=0$

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.

  $y^2+16=0$  ▸1^st equation

  $y-4=0$  ▸2^nd equation

  $y+4=0$  ▸3^rd equation

Solve the 1st equation $y^2+16=0$

• By subtracting 16 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 $\sqrt{-1}$

  $y^2+16-16=0-16$

  $y^2=-16$

  $\sqrt{y^2}=\sqrt{-16}$

  $\sqrt{y^2}=\sqrt{-1}\cdot \sqrt{16}$

  $y=\sqrt{-1}\cdot \pm 4$

  $y=\pm 4i$

Solving the 2nd equation $y-4=0$

• By adding 4 on both sides
• By simplifying the equation on both sides

  $y-4+4=0+4$

  $y=4$

Solving the 2nd equation $y+4=0$

• By subtracting 4 on both sides
• By simplifying the equation on both sides

  $y+4-4=0-4$

  $y=-4$

List the zeros of the functions given:

  $y=4i;y=-4i;y=4;y=-4$

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

So $y-4,y+4,y-4i$ & $y+4i$ are the factors of the function

Write the function $f\left(y\right)=y^4-256$ as product of linear factors:

  $f\left(y\right)=\left(y-4\right)\left(y+4\right)\left(y-4i\right)\left(y+4i\right)$

Conclusion:

The zeros of the given function $f\left(y\right)=y^4-256$ are: $y=4,-4,4i,\&-4i$

The function written as product of linear factors: $f\left(y\right)=\left(y-4\right)\left(y+4\right)\left(y-4i\right)\left(y+4i\right)$