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)
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