The zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i and the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

<h3>What are polynomial expressions?</h3>

Polynomial expressions are mathematical statements that are represented by variables, coefficients and operators

<h3>How to determine the zeros of the polynomial?</h3>

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Set the equation to 0

So, we have

(5y - 4)(5y + 4)(25y^2 + 16) = 0

Expand the equation

So, we have

5y - 4 = 0, 5y + 4 = 0 and 25y^2 + 16 = 0

This gives

5y = 4, 5y = -4 and 25y^2 = -16

Solve the factors of the equation

So, we have

y = 4/5, y = -4/5 and y = ±4/5√i

Hence, the zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i

How to write the polynomial as a product of the linear factors?

In (a), we have

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Hence, the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)