One factor of f(x) = 4x³ - 4x² - 16x + 16 is (x - 2). What are all the roots of the function? Use the Remainder Theorem.

Solution:

f(x) = 4x³ - 4x² - 16x + 16 [Given]

One factor is (x - 2)

By applying the remainder theorem, 

f(x) = (x-2)·q(x) + r(x), where q(x) = quotient polynomial and r(x) = remainder polynomial.

r(x) = 0 as (x-2) is a factor of f(x)

f(x) = (x-2)·q(x)

q(x) = f(x)/ (x-2)

[4x³ - 4x² - 16x + 16]/ (x - 2)

= [4x² (x - 1) - 16 (x - 1)]/ (x - 2)

By further simplification

= [(x - 1) (4x² - 16)]/ (x - 2)

Taking out 4 as common

= [(x - 1) 4 (x² - 4)]/ (x - 2)

Using the algebraic identity a² - b² = (a + b) (a - b)

= [(x - 1) 4 (x + 2) (x - 2)]/ (x - 2)

= [4 (x - 1) (x - 2) (x + 2)]/ (x - 2)

Therefore, the roots are 2, -2 and 1.

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One factor of f(x) = 4x³ - 4x² - 16x + 16 is (x - 2). What are all the roots of the function? Use the Remainder Theorem.

Summary:

One factor of f(x) = 4x³ - 4x² - 16x + 16 is (x - 2). All the roots of the function are 2, -2 and 1.